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如何创建目录的 zip 存档？

发布于 2024-08-13 09:22:34 字数 34 浏览 11 评论 0原文

如何在 Python 中创建目录结构的 zip 存档？

原文

How can I create a zip archive of a directory structure in Python?

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评论（30）

独行侠 2024-08-20 09:22:34

最简单的方法是使用 shutil.make_archive< /a>.它支持 zip 和 tar 格式。

import shutil
shutil.make_archive(output_filename, 'zip', dir_name)

如果您需要做比压缩整个目录更复杂的事情（例如跳过某些文件），那么您需要深入研究 zipfile 模块，正如其他人建议的那样。

The easiest way is to use shutil.make_archive. It supports both zip and tar formats.

import shutil
shutil.make_archive(output_filename, 'zip', dir_name)

If you need to do something more complicated than zipping the whole directory (such as skipping certain files), then you'll need to dig into the zipfile module as others have suggested.

回复 原文
因为看清所以看轻 2024-08-20 09:22:34

正如其他人指出的那样，您应该使用 zipfile。该文档告诉您哪些功能可用，但并没有真正解释如何使用它们来压缩整个目录。我认为用一些示例代码来解释是最简单的：

import os
import zipfile
    
def zipdir(path, ziph):
    # ziph is zipfile handle
    for root, dirs, files in os.walk(path):
        for file in files:
            ziph.write(os.path.join(root, file), 
                       os.path.relpath(os.path.join(root, file), 
                                       os.path.join(path, '..')))

with zipfile.ZipFile('Python.zip', 'w', zipfile.ZIP_DEFLATED) as zipf:
    zipdir('tmp/', zipf)

As others have pointed out, you should use zipfile. The documentation tells you what functions are available, but doesn't really explain how you can use them to zip an entire directory. I think it's easiest to explain with some example code:

import os
import zipfile
    
def zipdir(path, ziph):
    # ziph is zipfile handle
    for root, dirs, files in os.walk(path):
        for file in files:
            ziph.write(os.path.join(root, file), 
                       os.path.relpath(os.path.join(root, file), 
                                       os.path.join(path, '..')))

with zipfile.ZipFile('Python.zip', 'w', zipfile.ZIP_DEFLATED) as zipf:
    zipdir('tmp/', zipf)
回复 原文
找回味觉 2024-08-20 09:22:34

要将 mydirectory 的内容添加到新的 zip 文件，包括所有文件和子目录：

import os
import zipfile

zf = zipfile.ZipFile("myzipfile.zip", "w")
for dirname, subdirs, files in os.walk("mydirectory"):
    zf.write(dirname)
    for filename in files:
        zf.write(os.path.join(dirname, filename))
zf.close()

To add the contents of mydirectory to a new zip file, including all files and subdirectories:

import os
import zipfile

zf = zipfile.ZipFile("myzipfile.zip", "w")
for dirname, subdirs, files in os.walk("mydirectory"):
    zf.write(dirname)
    for filename in files:
        zf.write(os.path.join(dirname, filename))
zf.close()
回复 原文
夜声 2024-08-20 09:22:34

如何在 Python 中创建目录结构的 zip 存档？

在 Python 脚本中

在 Python 2.7+ 中，shutil 有一个 make_archive 函数。

from shutil import make_archive
make_archive(
  'zipfile_name', 
  'zip',           # the archive format - or tar, bztar, gztar 
  root_dir=None,   # root for archive - current working dir if None
  base_dir=None)   # start archiving from here - cwd if None too

这里，压缩档案将被命名为zipfile_name.zip。如果 base_dir 距离 root_dir 更远，它将排除不在 base_dir 中的文件，但仍将父目录中的文件存档到 <代码>root_dir。

我在使用 2.7 的 Cygwin 上测试时确实遇到了问题 - 它需要一个 root_dir 参数，用于 cwd：

make_archive('zipfile_name', 'zip', root_dir='.')

从 shell 使用 Python

您可以使用 shell 中的 Python 来执行此操作，也可以使用 zipfile 模块

$ python -m zipfile -c zipname sourcedir

： code>zipname 是您想要的目标文件的名称（如果需要，请添加 .zip，它不会自动执行），sourcedir 是目录的路径。

压缩 Python（或者只是不需要父目录）：

如果您尝试使用 __init__.py 和 __main__.py 压缩 python 包，并且您不需要父目录，它

$ python -m zipfile -c zipname sourcedir/*

会

$ python zipname

运行该包。 （请注意，您不能将子包作为压缩存档的入口点运行。）

压缩 Python 应用程序：

如果您有 python3.5+，并且特别想要压缩 Python 包，请使用 zipapp：

$ python -m zipapp myapp
$ python myapp.pyz

How can I create a zip archive of a directory structure in Python?

In a Python script

In Python 2.7+, shutil has a make_archive function.

from shutil import make_archive
make_archive(
  'zipfile_name', 
  'zip',           # the archive format - or tar, bztar, gztar 
  root_dir=None,   # root for archive - current working dir if None
  base_dir=None)   # start archiving from here - cwd if None too

Here the zipped archive will be named zipfile_name.zip. If base_dir is farther down from root_dir it will exclude files not in the base_dir, but still archive the files in the parent dirs up to the root_dir.

I did have an issue testing this on Cygwin with 2.7 - it wants a root_dir argument, for cwd:

make_archive('zipfile_name', 'zip', root_dir='.')

Using Python from the shell

You can do this with Python from the shell also using the zipfile module:

$ python -m zipfile -c zipname sourcedir

Where zipname is the name of the destination file you want (add .zip if you want it, it won't do it automatically) and sourcedir is the path to the directory.

Zipping up Python (or just don't want parent dir):

If you're trying to zip up a python package with a __init__.py and __main__.py, and you don't want the parent dir, it's

$ python -m zipfile -c zipname sourcedir/*

And

$ python zipname

would run the package. (Note that you can't run subpackages as the entry point from a zipped archive.)

Zipping a Python app:

If you have python3.5+, and specifically want to zip up a Python package, use zipapp:

$ python -m zipapp myapp
$ python myapp.pyz
回复 原文
穿透光 2024-08-20 09:22:34

此函数将递归地压缩目录树，压缩文件，并在存档中记录正确的相对文件名。归档条目与 zip -r output.zip source_dir 生成的条目相同。

import os
import zipfile
def make_zipfile(output_filename, source_dir):
    relroot = os.path.abspath(os.path.join(source_dir, os.pardir))
    with zipfile.ZipFile(output_filename, "w", zipfile.ZIP_DEFLATED) as zip:
        for root, dirs, files in os.walk(source_dir):
            # add directory (needed for empty dirs)
            zip.write(root, os.path.relpath(root, relroot))
            for file in files:
                filename = os.path.join(root, file)
                if os.path.isfile(filename): # regular files only
                    arcname = os.path.join(os.path.relpath(root, relroot), file)
                    zip.write(filename, arcname)

This function will recursively zip up a directory tree, compressing the files, and recording the correct relative filenames in the archive. The archive entries are the same as those generated by zip -r output.zip source_dir.

import os
import zipfile
def make_zipfile(output_filename, source_dir):
    relroot = os.path.abspath(os.path.join(source_dir, os.pardir))
    with zipfile.ZipFile(output_filename, "w", zipfile.ZIP_DEFLATED) as zip:
        for root, dirs, files in os.walk(source_dir):
            # add directory (needed for empty dirs)
            zip.write(root, os.path.relpath(root, relroot))
            for file in files:
                filename = os.path.join(root, file)
                if os.path.isfile(filename): # regular files only
                    arcname = os.path.join(os.path.relpath(root, relroot), file)
                    zip.write(filename, arcname)
回复 原文
少女七分熟 2024-08-20 09:22:34

使用 python 3.9，pathlib & zipfile 模块您可以从系统中的任何位置创建 zip 文件。

def zip_dir(dir: Union[Path, str], filename: Union[Path, str]):
    """Zip the provided directory without navigating to that directory using `pathlib` module"""

    # Convert to Path object
    dir = Path(dir)

    with zipfile.ZipFile(filename, "w", zipfile.ZIP_DEFLATED) as zip_file:
        for entry in dir.rglob("*"):
            zip_file.write(entry, entry.relative_to(dir))

它很整洁，有打字功能，并且代码较少。

With python 3.9, pathlib & zipfile module you can create a zip files from anywhere in the system.

def zip_dir(dir: Union[Path, str], filename: Union[Path, str]):
    """Zip the provided directory without navigating to that directory using `pathlib` module"""

    # Convert to Path object
    dir = Path(dir)

    with zipfile.ZipFile(filename, "w", zipfile.ZIP_DEFLATED) as zip_file:
        for entry in dir.rglob("*"):
            zip_file.write(entry, entry.relative_to(dir))

It is neat, typed, and has less code.

回复 原文
给不了的爱 2024-08-20 09:22:34

使用 pathlib 模块的现代 Python (3.6+)用于简洁的类似 OOP 的路径处理，以及 pathlib。 Path.rglob() 用于递归通配符。据我所知，这相当于 George V. Reilly 的答案：压缩 zip，最上面的元素是一个目录，保留空目录，使用相对路径。

from pathlib import Path
from zipfile import ZIP_DEFLATED, ZipFile

from os import PathLike
from typing import Union


def zip_dir(zip_name: str, source_dir: Union[str, PathLike]):
    src_path = Path(source_dir).expanduser().resolve(strict=True)
    with ZipFile(zip_name, 'w', ZIP_DEFLATED) as zf:
        for file in src_path.rglob('*'):
            zf.write(file, file.relative_to(src_path.parent))

注意：正如可选类型提示所示， zip_name 不能是 Path 对象（将被修复在 3.6.2+ 中）。

Modern Python (3.6+) using the pathlib module for concise OOP-like handling of paths, and pathlib.Path.rglob() for recursive globbing. As far as I can tell, this is equivalent to George V. Reilly's answer: zips with compression, the topmost element is a directory, keeps empty dirs, uses relative paths.

from pathlib import Path
from zipfile import ZIP_DEFLATED, ZipFile

from os import PathLike
from typing import Union


def zip_dir(zip_name: str, source_dir: Union[str, PathLike]):
    src_path = Path(source_dir).expanduser().resolve(strict=True)
    with ZipFile(zip_name, 'w', ZIP_DEFLATED) as zf:
        for file in src_path.rglob('*'):
            zf.write(file, file.relative_to(src_path.parent))

Note: as optional type hints indicate, zip_name can't be a Path object (would be fixed in 3.6.2+).

回复 原文
北斗星光 2024-08-20 09:22:34

使用shutil，它是python标准库集的一部分。
使用 Shutil 非常简单（参见下面的代码）：

  • 第一个参数：生成的 zip/tar 文件的文件名，
  • 第二个参数：zip/tar，
  • 第三个参数：dir_name

代码：

import shutil
shutil.make_archive('/home/user/Desktop/Filename','zip','/home/username/Desktop/Directory')

Use shutil, which is part of python standard library set.
Using shutil is so simple(see code below):

  • 1st arg: Filename of resultant zip/tar file,
  • 2nd arg: zip/tar,
  • 3rd arg: dir_name

Code:

import shutil
shutil.make_archive('/home/user/Desktop/Filename','zip','/home/username/Desktop/Directory')
回复 原文
情愿 2024-08-20 09:22:34

要向生成的 zip 文件添加压缩，请查看 这个链接。

您需要更改

zip = zipfile.ZipFile('Python.zip', 'w')

：

zip = zipfile.ZipFile('Python.zip', 'w', zipfile.ZIP_DEFLATED)

For adding compression to the resulting zip file, check out this link.

You need to change:

zip = zipfile.ZipFile('Python.zip', 'w')

to

zip = zipfile.ZipFile('Python.zip', 'w', zipfile.ZIP_DEFLATED)
回复 原文
音栖息无 2024-08-20 09:22:34

要保留要归档的父目录下的文件夹层次结构：

from pathlib import Path
import zipfile

fp_zip = Path("output.zip")
path_to_archive = Path("./path-to-archive")

with zipfile.ZipFile(fp_zip, "w", zipfile.ZIP_DEFLATED) as zipf:
    for fp in path_to_archive.glob("**/*"):
        zipf.write(fp, arcname=fp.relative_to(path_to_archive))

To retain the folder hierarchy under the parent directory to be archived:

from pathlib import Path
import zipfile

fp_zip = Path("output.zip")
path_to_archive = Path("./path-to-archive")

with zipfile.ZipFile(fp_zip, "w", zipfile.ZIP_DEFLATED) as zipf:
    for fp in path_to_archive.glob("**/*"):
        zipf.write(fp, arcname=fp.relative_to(path_to_archive))
回复 原文
心凉 2024-08-20 09:22:34

我对 Mark Byers 提供的代码进行了一些更改。如果有的话，下面的函数还将添加空目录。示例应该可以更清楚地说明添加到 zip 中的路径是什么。

#!/usr/bin/env python
import os
import zipfile

def addDirToZip(zipHandle, path, basePath=""):
    """
    Adding directory given by \a path to opened zip file \a zipHandle

    @param basePath path that will be removed from \a path when adding to archive

    Examples:
        # add whole "dir" to "test.zip" (when you open "test.zip" you will see only "dir")
        zipHandle = zipfile.ZipFile('test.zip', 'w')
        addDirToZip(zipHandle, 'dir')
        zipHandle.close()

        # add contents of "dir" to "test.zip" (when you open "test.zip" you will see only it's contents)
        zipHandle = zipfile.ZipFile('test.zip', 'w')
        addDirToZip(zipHandle, 'dir', 'dir')
        zipHandle.close()

        # add contents of "dir/subdir" to "test.zip" (when you open "test.zip" you will see only contents of "subdir")
        zipHandle = zipfile.ZipFile('test.zip', 'w')
        addDirToZip(zipHandle, 'dir/subdir', 'dir/subdir')
        zipHandle.close()

        # add whole "dir/subdir" to "test.zip" (when you open "test.zip" you will see only "subdir")
        zipHandle = zipfile.ZipFile('test.zip', 'w')
        addDirToZip(zipHandle, 'dir/subdir', 'dir')
        zipHandle.close()

        # add whole "dir/subdir" with full path to "test.zip" (when you open "test.zip" you will see only "dir" and inside it only "subdir")
        zipHandle = zipfile.ZipFile('test.zip', 'w')
        addDirToZip(zipHandle, 'dir/subdir')
        zipHandle.close()

        # add whole "dir" and "otherDir" (with full path) to "test.zip" (when you open "test.zip" you will see only "dir" and "otherDir")
        zipHandle = zipfile.ZipFile('test.zip', 'w')
        addDirToZip(zipHandle, 'dir')
        addDirToZip(zipHandle, 'otherDir')
        zipHandle.close()
    """
    basePath = basePath.rstrip("\\/") + ""
    basePath = basePath.rstrip("\\/")
    for root, dirs, files in os.walk(path):
        # add dir itself (needed for empty dirs
        zipHandle.write(os.path.join(root, "."))
        # add files
        for file in files:
            filePath = os.path.join(root, file)
            inZipPath = filePath.replace(basePath, "", 1).lstrip("\\/")
            #print filePath + " , " + inZipPath
            zipHandle.write(filePath, inZipPath)

上面是一个简单的函数，应该适用于简单的情况。你可以在我的要点中找到更优雅的课程：
https://gist.github.com/Eccenux/17526123107ca0ac28e6

I've made some changes to code given by Mark Byers. Below function will also adds empty directories if you have them. Examples should make it more clear what is the path added to the zip.

#!/usr/bin/env python
import os
import zipfile

def addDirToZip(zipHandle, path, basePath=""):
    """
    Adding directory given by \a path to opened zip file \a zipHandle

    @param basePath path that will be removed from \a path when adding to archive

    Examples:
        # add whole "dir" to "test.zip" (when you open "test.zip" you will see only "dir")
        zipHandle = zipfile.ZipFile('test.zip', 'w')
        addDirToZip(zipHandle, 'dir')
        zipHandle.close()

        # add contents of "dir" to "test.zip" (when you open "test.zip" you will see only it's contents)
        zipHandle = zipfile.ZipFile('test.zip', 'w')
        addDirToZip(zipHandle, 'dir', 'dir')
        zipHandle.close()

        # add contents of "dir/subdir" to "test.zip" (when you open "test.zip" you will see only contents of "subdir")
        zipHandle = zipfile.ZipFile('test.zip', 'w')
        addDirToZip(zipHandle, 'dir/subdir', 'dir/subdir')
        zipHandle.close()

        # add whole "dir/subdir" to "test.zip" (when you open "test.zip" you will see only "subdir")
        zipHandle = zipfile.ZipFile('test.zip', 'w')
        addDirToZip(zipHandle, 'dir/subdir', 'dir')
        zipHandle.close()

        # add whole "dir/subdir" with full path to "test.zip" (when you open "test.zip" you will see only "dir" and inside it only "subdir")
        zipHandle = zipfile.ZipFile('test.zip', 'w')
        addDirToZip(zipHandle, 'dir/subdir')
        zipHandle.close()

        # add whole "dir" and "otherDir" (with full path) to "test.zip" (when you open "test.zip" you will see only "dir" and "otherDir")
        zipHandle = zipfile.ZipFile('test.zip', 'w')
        addDirToZip(zipHandle, 'dir')
        addDirToZip(zipHandle, 'otherDir')
        zipHandle.close()
    """
    basePath = basePath.rstrip("\\/") + ""
    basePath = basePath.rstrip("\\/")
    for root, dirs, files in os.walk(path):
        # add dir itself (needed for empty dirs
        zipHandle.write(os.path.join(root, "."))
        # add files
        for file in files:
            filePath = os.path.join(root, file)
            inZipPath = filePath.replace(basePath, "", 1).lstrip("\\/")
            #print filePath + " , " + inZipPath
            zipHandle.write(filePath, inZipPath)

Above is a simple function that should work for simple cases. You can find more elegant class in my Gist:
https://gist.github.com/Eccenux/17526123107ca0ac28e6

回复 原文
雨后彩虹 2024-08-20 09:22:34

我有另一个代码示例可能会有所帮助，使用 python3、pathlib 和 zipfile。
它应该可以在任何操作系统中工作。

from pathlib import Path
import zipfile
from datetime import datetime

DATE_FORMAT = '%y%m%d'


def date_str():
    """returns the today string year, month, day"""
    return '{}'.format(datetime.now().strftime(DATE_FORMAT))


def zip_name(path):
    """returns the zip filename as string"""
    cur_dir = Path(path).resolve()
    parent_dir = cur_dir.parents[0]
    zip_filename = '{}/{}_{}.zip'.format(parent_dir, cur_dir.name, date_str())
    p_zip = Path(zip_filename)
    n = 1
    while p_zip.exists():
        zip_filename = ('{}/{}_{}_{}.zip'.format(parent_dir, cur_dir.name,
                                             date_str(), n))
        p_zip = Path(zip_filename)
        n += 1
    return zip_filename


def all_files(path):
    """iterator returns all files and folders from path as absolute path string
    """
    for child in Path(path).iterdir():
        yield str(child)
        if child.is_dir():
            for grand_child in all_files(str(child)):
                yield str(Path(grand_child))


def zip_dir(path):
    """generate a zip"""
    zip_filename = zip_name(path)
    zip_file = zipfile.ZipFile(zip_filename, 'w')
    print('create:', zip_filename)
    for file in all_files(path):
        print('adding... ', file)
        zip_file.write(file)
    zip_file.close()


if __name__ == '__main__':
    zip_dir('.')
    print('end!')

I have another code example that may help, using python3, pathlib and zipfile.
It should work in any OS.

from pathlib import Path
import zipfile
from datetime import datetime

DATE_FORMAT = '%y%m%d'


def date_str():
    """returns the today string year, month, day"""
    return '{}'.format(datetime.now().strftime(DATE_FORMAT))


def zip_name(path):
    """returns the zip filename as string"""
    cur_dir = Path(path).resolve()
    parent_dir = cur_dir.parents[0]
    zip_filename = '{}/{}_{}.zip'.format(parent_dir, cur_dir.name, date_str())
    p_zip = Path(zip_filename)
    n = 1
    while p_zip.exists():
        zip_filename = ('{}/{}_{}_{}.zip'.format(parent_dir, cur_dir.name,
                                             date_str(), n))
        p_zip = Path(zip_filename)
        n += 1
    return zip_filename


def all_files(path):
    """iterator returns all files and folders from path as absolute path string
    """
    for child in Path(path).iterdir():
        yield str(child)
        if child.is_dir():
            for grand_child in all_files(str(child)):
                yield str(Path(grand_child))


def zip_dir(path):
    """generate a zip"""
    zip_filename = zip_name(path)
    zip_file = zipfile.ZipFile(zip_filename, 'w')
    print('create:', zip_filename)
    for file in all_files(path):
        print('adding... ', file)
        zip_file.write(file)
    zip_file.close()


if __name__ == '__main__':
    zip_dir('.')
    print('end!')
回复 原文
书信已泛黄 2024-08-20 09:22:34

如果您想要像任何常见图形文件管理器的压缩文件夹一样的功能，您可以使用以下代码，它使用 zipfile 模块。使用此代码，您将获得以路径作为根文件夹的 zip 文件。

import os
import zipfile

def zipdir(path, ziph):
    # Iterate all the directories and files
    for root, dirs, files in os.walk(path):
        # Create a prefix variable with the folder structure inside the path folder. 
        # So if a file is at the path directory will be at the root directory of the zip file
        # so the prefix will be empty. If the file belongs to a containing folder of path folder 
        # then the prefix will be that folder.
        if root.replace(path,'') == '':
                prefix = ''
        else:
                # Keep the folder structure after the path folder, append a '/' at the end 
                # and remome the first character, if it is a '/' in order to have a path like 
                # folder1/folder2/file.txt
                prefix = root.replace(path, '') + '/'
                if (prefix[0] == '/'):
                        prefix = prefix[1:]
        for filename in files:
                actual_file_path = root + '/' + filename
                zipped_file_path = prefix + filename
                zipf.write( actual_file_path, zipped_file_path)


zipf = zipfile.ZipFile('Python.zip', 'w', zipfile.ZIP_DEFLATED)
zipdir('/tmp/justtest/', zipf)
zipf.close()

If you want a functionality like the compress folder of any common graphical file manager you can use the following code, it uses the zipfile module. Using this code you will have the zip file with the path as its root folder.

import os
import zipfile

def zipdir(path, ziph):
    # Iterate all the directories and files
    for root, dirs, files in os.walk(path):
        # Create a prefix variable with the folder structure inside the path folder. 
        # So if a file is at the path directory will be at the root directory of the zip file
        # so the prefix will be empty. If the file belongs to a containing folder of path folder 
        # then the prefix will be that folder.
        if root.replace(path,'') == '':
                prefix = ''
        else:
                # Keep the folder structure after the path folder, append a '/' at the end 
                # and remome the first character, if it is a '/' in order to have a path like 
                # folder1/folder2/file.txt
                prefix = root.replace(path, '') + '/'
                if (prefix[0] == '/'):
                        prefix = prefix[1:]
        for filename in files:
                actual_file_path = root + '/' + filename
                zipped_file_path = prefix + filename
                zipf.write( actual_file_path, zipped_file_path)


zipf = zipfile.ZipFile('Python.zip', 'w', zipfile.ZIP_DEFLATED)
zipdir('/tmp/justtest/', zipf)
zipf.close()
回复 原文
行雁书 2024-08-20 09:22:34

这里有很多答案，我希望我可以贡献自己的版本，它基于原始答案（顺便说一下），但具有更图形化的视角，还为每个 zipfile 设置使用上下文并对 os.walk() 进行排序，以获得有序的输出。

有了这些文件夹和文件（以及其他文件夹），我想为每个 cap_ 文件夹创建一个 .zip：

$ tree -d
.
├── cap_01
|    ├── 0101000001.json
|    ├── 0101000002.json
|    ├── 0101000003.json
|
├── cap_02
|    ├── 0201000001.json
|    ├── 0201000002.json
|    ├── 0201001003.json
|
├── cap_03
|    ├── 0301000001.json
|    ├── 0301000002.json
|    ├── 0301000003.json
| 
├── docs
|    ├── map.txt
|    ├── main_data.xml
|
├── core_files
     ├── core_master
     ├── core_slave

这是我应用的内容，其中包含注释，以便更好地理解过程。

$ cat zip_cap_dirs.py 
""" Zip 'cap_*' directories. """           
import os                                                                       
import zipfile as zf                                                            


for root, dirs, files in sorted(os.walk('.')):                                                                                               
    if 'cap_' in root:                                                          
        print(f"Compressing: {root}")                                           
        # Defining .zip name, according to Capítulo.                            
        cap_dir_zip = '{}.zip'.format(root)                                     
        # Opening zipfile context for current root dir.                         
        with zf.ZipFile(cap_dir_zip, 'w', zf.ZIP_DEFLATED) as new_zip:          
            # Iterating over os.walk list of files for the current root dir.    
            for f in files:                                                     
                # Defining relative path to files from current root dir.        
                f_path = os.path.join(root, f)                                  
                # Writing the file on the .zip file of the context              
                new_zip.write(f_path)

基本上，对于 os.walk(path) 的每次迭代，我都会打开一个用于 zipfile 设置的上下文，然后迭代 files ，它是 root 目录中的文件的 list，根据当前 root 目录形成每个文件的相对路径，附加到 < code>zipfile 正在运行的上下文。

输出如下所示：

$ python3 zip_cap_dirs.py
Compressing: ./cap_01
Compressing: ./cap_02
Compressing: ./cap_03

要查看每个 .zip 目录的内容，可以使用 less 命令：

$ less cap_01.zip

Archive:  cap_01.zip
 Length   Method    Size  Cmpr    Date    Time   CRC-32   Name
--------  ------  ------- ---- ---------- ----- --------  ----
  22017  Defl:N     2471  89% 2019-09-05 08:05 7a3b5ec6  cap_01/0101000001.json
  21998  Defl:N     2471  89% 2019-09-05 08:05 155bece7  cap_01/0101000002.json
  23236  Defl:N     2573  89% 2019-09-05 08:05 55fced20  cap_01/0101000003.json
--------          ------- ---                           -------
  67251             7515  89%                            3 files

So many answers here, and I hope I might contribute with my own version, which is based on the original answer (by the way), but with a more graphical perspective, also using context for each zipfile setup and sorting os.walk(), in order to have a ordered output.

Having these folders and them files (among other folders), I wanted to create a .zip for each cap_ folder:

$ tree -d
.
├── cap_01
|    ├── 0101000001.json
|    ├── 0101000002.json
|    ├── 0101000003.json
|
├── cap_02
|    ├── 0201000001.json
|    ├── 0201000002.json
|    ├── 0201001003.json
|
├── cap_03
|    ├── 0301000001.json
|    ├── 0301000002.json
|    ├── 0301000003.json
| 
├── docs
|    ├── map.txt
|    ├── main_data.xml
|
├── core_files
     ├── core_master
     ├── core_slave

Here's what I applied, with comments for better understanding of the process.

$ cat zip_cap_dirs.py 
""" Zip 'cap_*' directories. """           
import os                                                                       
import zipfile as zf                                                            


for root, dirs, files in sorted(os.walk('.')):                                                                                               
    if 'cap_' in root:                                                          
        print(f"Compressing: {root}")                                           
        # Defining .zip name, according to Capítulo.                            
        cap_dir_zip = '{}.zip'.format(root)                                     
        # Opening zipfile context for current root dir.                         
        with zf.ZipFile(cap_dir_zip, 'w', zf.ZIP_DEFLATED) as new_zip:          
            # Iterating over os.walk list of files for the current root dir.    
            for f in files:                                                     
                # Defining relative path to files from current root dir.        
                f_path = os.path.join(root, f)                                  
                # Writing the file on the .zip file of the context              
                new_zip.write(f_path)

Basically, for each iteration over os.walk(path), I'm opening a context for zipfile setup and afterwards, iterating iterating over files, which is a list of files from root directory, forming the relative path for each file based on the current root directory, appending to the zipfile context which is running.

And the output is presented like this:

$ python3 zip_cap_dirs.py
Compressing: ./cap_01
Compressing: ./cap_02
Compressing: ./cap_03

To see the contents of each .zip directory, you can use less command:

$ less cap_01.zip

Archive:  cap_01.zip
 Length   Method    Size  Cmpr    Date    Time   CRC-32   Name
--------  ------  ------- ---- ---------- ----- --------  ----
  22017  Defl:N     2471  89% 2019-09-05 08:05 7a3b5ec6  cap_01/0101000001.json
  21998  Defl:N     2471  89% 2019-09-05 08:05 155bece7  cap_01/0101000002.json
  23236  Defl:N     2573  89% 2019-09-05 08:05 55fced20  cap_01/0101000003.json
--------          ------- ---                           -------
  67251             7515  89%                            3 files
回复 原文
小苏打饼 2024-08-20 09:22:34

使用 pathlib.Path 的解决方案，它独立于所使用的操作系统：

import zipfile
from pathlib import Path

def zip_dir(path: Path, zip_file_path: Path):
    """Zip all contents of path to zip_file"""
    files_to_zip = [
        file for file in path.glob('*') if file.is_file()]
    with zipfile.ZipFile(
        zip_file_path, 'w', zipfile.ZIP_DEFLATED) as zip_f:
        for file in files_to_zip:
            print(file.name)
            zip_f.write(file, file.name)

current_dir = Path.cwd()  
tozip_dir = current_dir / "test"
zip_dir(
    tozip_dir, current_dir / 'dir.zip')

A solution using pathlib.Path, which is independent of the OS used:

import zipfile
from pathlib import Path

def zip_dir(path: Path, zip_file_path: Path):
    """Zip all contents of path to zip_file"""
    files_to_zip = [
        file for file in path.glob('*') if file.is_file()]
    with zipfile.ZipFile(
        zip_file_path, 'w', zipfile.ZIP_DEFLATED) as zip_f:
        for file in files_to_zip:
            print(file.name)
            zip_f.write(file, file.name)

current_dir = Path.cwd()  
tozip_dir = current_dir / "test"
zip_dir(
    tozip_dir, current_dir / 'dir.zip')
回复 原文
羁绊已千年 2024-08-20 09:22:34

为了提供更大的灵活性，例如按名称选择目录/文件，请使用：

import os
import zipfile

def zipall(ob, path, rel=""):
    basename = os.path.basename(path)
    if os.path.isdir(path):
        if rel == "":
            rel = basename
        ob.write(path, os.path.join(rel))
        for root, dirs, files in os.walk(path):
            for d in dirs:
                zipall(ob, os.path.join(root, d), os.path.join(rel, d))
            for f in files:
                ob.write(os.path.join(root, f), os.path.join(rel, f))
            break
    elif os.path.isfile(path):
        ob.write(path, os.path.join(rel, basename))
    else:
        pass

对于文件树：

.
├── dir
│   ├── dir2
│   │   └── file2.txt
│   ├── dir3
│   │   └── file3.txt
│   └── file.txt
├── dir4
│   ├── dir5
│   └── file4.txt
├── listdir.zip
├── main.py
├── root.txt
└── selective.zip

您可以例如仅选择 dir4 和 root.txt：

cwd = os.getcwd()
files = [os.path.join(cwd, f) for f in ['dir4', 'root.txt']]

with zipfile.ZipFile("selective.zip", "w" ) as myzip:
    for f in files:
        zipall(myzip, f)

或者仅选择 listdir 在脚本调用目录中并从那里添加所有内容：

with zipfile.ZipFile("listdir.zip", "w" ) as myzip:
    for f in os.listdir():
        if f == "listdir.zip":
            # Creating a listdir.zip in the same directory
            # will include listdir.zip inside itself, beware of this
            continue
        zipall(myzip, f)

To give more flexibility, e.g. select directory/file by name use:

import os
import zipfile

def zipall(ob, path, rel=""):
    basename = os.path.basename(path)
    if os.path.isdir(path):
        if rel == "":
            rel = basename
        ob.write(path, os.path.join(rel))
        for root, dirs, files in os.walk(path):
            for d in dirs:
                zipall(ob, os.path.join(root, d), os.path.join(rel, d))
            for f in files:
                ob.write(os.path.join(root, f), os.path.join(rel, f))
            break
    elif os.path.isfile(path):
        ob.write(path, os.path.join(rel, basename))
    else:
        pass

For a file tree:

.
├── dir
│   ├── dir2
│   │   └── file2.txt
│   ├── dir3
│   │   └── file3.txt
│   └── file.txt
├── dir4
│   ├── dir5
│   └── file4.txt
├── listdir.zip
├── main.py
├── root.txt
└── selective.zip

You can e.g. select only dir4 and root.txt:

cwd = os.getcwd()
files = [os.path.join(cwd, f) for f in ['dir4', 'root.txt']]

with zipfile.ZipFile("selective.zip", "w" ) as myzip:
    for f in files:
        zipall(myzip, f)

Or just listdir in script invocation directory and add everything from there:

with zipfile.ZipFile("listdir.zip", "w" ) as myzip:
    for f in os.listdir():
        if f == "listdir.zip":
            # Creating a listdir.zip in the same directory
            # will include listdir.zip inside itself, beware of this
            continue
        zipall(myzip, f)
回复 原文
装纯掩盖桑 2024-08-20 09:22:34

使用 shutil 时请注意，您应该在 base_name arg 中包含输出目录路径：

import shutil


shutil.make_archive(
  base_name=output_dir_path + output_filename_without_extension, 
  format="zip", 
  root_dir=input_root_dir)

While using shutil note that you should include output directory path in base_name arg:

import shutil


shutil.make_archive(
  base_name=output_dir_path + output_filename_without_extension, 
  format="zip", 
  root_dir=input_root_dir)
回复 原文
甚是思念 2024-08-20 09:22:34

您可能想查看 zipfile 模块；有文档 http://docs.python.org/library/zipfile.html 。

您可能还需要 os.walk() 来索引目录结构。

You probably want to look at the zipfile module; there's documentation at http://docs.python.org/library/zipfile.html.

You may also want os.walk() to index the directory structure.

回复 原文
巨坚强 2024-08-20 09:22:34

这是 Nux 给出的答案的一个变体，对我有用：

def WriteDirectoryToZipFile( zipHandle, srcPath, zipLocalPath = "", zipOperation = zipfile.ZIP_DEFLATED ):
    basePath = os.path.split( srcPath )[ 0 ]
    for root, dirs, files in os.walk( srcPath ):
        p = os.path.join( zipLocalPath, root [ ( len( basePath ) + 1 ) : ] )
        # add dir
        zipHandle.write( root, p, zipOperation )
        # add files
        for f in files:
            filePath = os.path.join( root, f )
            fileInZipPath = os.path.join( p, f )
            zipHandle.write( filePath, fileInZipPath, zipOperation )

Here is a variation on the answer given by Nux that works for me:

def WriteDirectoryToZipFile( zipHandle, srcPath, zipLocalPath = "", zipOperation = zipfile.ZIP_DEFLATED ):
    basePath = os.path.split( srcPath )[ 0 ]
    for root, dirs, files in os.walk( srcPath ):
        p = os.path.join( zipLocalPath, root [ ( len( basePath ) + 1 ) : ] )
        # add dir
        zipHandle.write( root, p, zipOperation )
        # add files
        for f in files:
            filePath = os.path.join( root, f )
            fileInZipPath = os.path.join( p, f )
            zipHandle.write( filePath, fileInZipPath, zipOperation )
回复 原文
短暂陪伴 2024-08-20 09:22:34

尝试下面的方法。它对我有用。

import zipfile, os
zipf = "compress.zip"  
def main():
    directory = r"Filepath"
    toZip(directory)
def toZip(directory):
    zippedHelp = zipfile.ZipFile(zipf, "w", compression=zipfile.ZIP_DEFLATED )

    list = os.listdir(directory)
    for file_list in list:
        file_name = os.path.join(directory,file_list)

        if os.path.isfile(file_name):
            print file_name
            zippedHelp.write(file_name)
        else:
            addFolderToZip(zippedHelp,file_list,directory)
            print "---------------Directory Found-----------------------"
    zippedHelp.close()

def addFolderToZip(zippedHelp,folder,directory):
    path=os.path.join(directory,folder)
    print path
    file_list=os.listdir(path)
    for file_name in file_list:
        file_path=os.path.join(path,file_name)
        if os.path.isfile(file_path):
            zippedHelp.write(file_path)
        elif os.path.isdir(file_name):
            print "------------------sub directory found--------------------"
            addFolderToZip(zippedHelp,file_name,path)


if __name__=="__main__":
    main()

Try the below one .it worked for me.

import zipfile, os
zipf = "compress.zip"  
def main():
    directory = r"Filepath"
    toZip(directory)
def toZip(directory):
    zippedHelp = zipfile.ZipFile(zipf, "w", compression=zipfile.ZIP_DEFLATED )

    list = os.listdir(directory)
    for file_list in list:
        file_name = os.path.join(directory,file_list)

        if os.path.isfile(file_name):
            print file_name
            zippedHelp.write(file_name)
        else:
            addFolderToZip(zippedHelp,file_list,directory)
            print "---------------Directory Found-----------------------"
    zippedHelp.close()

def addFolderToZip(zippedHelp,folder,directory):
    path=os.path.join(directory,folder)
    print path
    file_list=os.listdir(path)
    for file_name in file_list:
        file_path=os.path.join(path,file_name)
        if os.path.isfile(file_path):
            zippedHelp.write(file_path)
        elif os.path.isdir(file_name):
            print "------------------sub directory found--------------------"
            addFolderToZip(zippedHelp,file_name,path)


if __name__=="__main__":
    main()
回复 原文
↙温凉少女 2024-08-20 09:22:34

假设您要压缩当前目录中的所有文件夹（子目录）。

for root, dirs, files in os.walk("."):
    for sub_dir in dirs:
        zip_you_want = sub_dir+".zip"
        zip_process = zipfile.ZipFile(zip_you_want, "w", zipfile.ZIP_DEFLATED)
        zip_process.write(file_you_want_to_include)
        zip_process.close()

        print("Successfully zipped directory: {sub_dir}".format(sub_dir=sub_dir))

Say you want to Zip all the folders(sub directories) in the current directory.

for root, dirs, files in os.walk("."):
    for sub_dir in dirs:
        zip_you_want = sub_dir+".zip"
        zip_process = zipfile.ZipFile(zip_you_want, "w", zipfile.ZIP_DEFLATED)
        zip_process.write(file_you_want_to_include)
        zip_process.close()

        print("Successfully zipped directory: {sub_dir}".format(sub_dir=sub_dir))
回复 原文
执着的年纪 2024-08-20 09:22:34

压缩文件或树（目录及其子目录）。

from pathlib import Path
from zipfile import ZipFile, ZIP_DEFLATED

def make_zip(tree_path, zip_path, mode='w', skip_empty_dir=False):
    with ZipFile(zip_path, mode=mode, compression=ZIP_DEFLATED) as zf:
        paths = [Path(tree_path)]
        while paths:
            p = paths.pop()
            if p.is_dir():
                paths.extend(p.iterdir())
                if skip_empty_dir:
                    continue
            zf.write(p)

要附加到现有存档，请传递 mode='a'，以创建新的存档 mode='w'（上面的默认值）。假设您想将 3 个不同的目录树捆绑在同一个存档下。

make_zip(path_to_tree1, path_to_arch, mode='w')
make_zip(path_to_tree2, path_to_arch, mode='a')
make_zip(path_to_file3, path_to_arch, mode='a')

Zip a file or a tree (a directory and its sub-directories).

from pathlib import Path
from zipfile import ZipFile, ZIP_DEFLATED

def make_zip(tree_path, zip_path, mode='w', skip_empty_dir=False):
    with ZipFile(zip_path, mode=mode, compression=ZIP_DEFLATED) as zf:
        paths = [Path(tree_path)]
        while paths:
            p = paths.pop()
            if p.is_dir():
                paths.extend(p.iterdir())
                if skip_empty_dir:
                    continue
            zf.write(p)

To append to an existing archive, pass mode='a', to create a fresh archive mode='w' (the default in the above). So let's say you want to bundle 3 different directory trees under the same archive.

make_zip(path_to_tree1, path_to_arch, mode='w')
make_zip(path_to_tree2, path_to_arch, mode='a')
make_zip(path_to_file3, path_to_arch, mode='a')
回复 原文
人疚 2024-08-20 09:22:34

显而易见的方法是使用 Shutil，就像第二个最佳答案所说的那样，但是如果您出于某种原因仍然希望使用 ZipFile，并且如果您在这样做时遇到一些麻烦（例如 Windows 中的 ERR 13 等） ，您可以使用此修复：

import os
import zipfile

def retrieve_file_paths(dirName):
  filePaths = []
  for root, directories, files in os.walk(dirName):
    for filename in files:
        filePath = os.path.join(root, filename)
        filePaths.append(filePath)
  return filePaths
 
def main(dir_name, output_filename):
  filePaths = retrieve_file_paths(dir_name)
   
  zip_file = zipfile.ZipFile(output_filename+'.zip', 'w')
  with zip_file:
    for file in filePaths:
      zip_file.write(file)

main("my_dir", "my_dir_archived")

此修复会递归地迭代给定文件夹中的每个子文件夹/文件，并将它们写入 zip 文件，而不是尝试直接压缩文件夹。

The obvious way to go would be to go with shutil, Like the second top answer says so, But if you still wish to go with ZipFile for some reason, And if you are getting some trouble doing that (Like ERR 13 in Windows etc), You can use this fix:

import os
import zipfile

def retrieve_file_paths(dirName):
  filePaths = []
  for root, directories, files in os.walk(dirName):
    for filename in files:
        filePath = os.path.join(root, filename)
        filePaths.append(filePath)
  return filePaths
 
def main(dir_name, output_filename):
  filePaths = retrieve_file_paths(dir_name)
   
  zip_file = zipfile.ZipFile(output_filename+'.zip', 'w')
  with zip_file:
    for file in filePaths:
      zip_file.write(file)

main("my_dir", "my_dir_archived")

This one recursively iterates through every sub-folder/file in your given folder, And writes them to a zip file instead of attempting to directly zip a folder.

回复 原文
嘴硬脾气大 2024-08-20 09:22:34

之前的答案完全忽略了一件事情，即当您在 Windows 上运行代码时，使用 os.path.join() 可以轻松返回 POSIX 不兼容的路径。当使用 Linux 上的任何常见存档软件处理生成的存档时，生成的存档将包含名称中带有文字反斜杠的文件，这不是您想要的。请使用 path.as_posix() 作为 arcname 参数！

import zipfile
from pathlib import Path
with zipfile.ZipFile("archive.zip", "w", zipfile.ZIP_DEFLATED) as zf:
    for path in Path("include_all_of_this_folder").rglob("*"):
        zf.write(path, path.as_posix())

One thing completely missed by previous answers is the fact that using os.path.join() can easily return POSIX-incompatible paths when you run the code on Windows. The resulting archive will contain files with literal backslashes in their names when processing it with any common archive software on Linux, which is not what you want. Use path.as_posix() for the arcname parameter instead!

import zipfile
from pathlib import Path
with zipfile.ZipFile("archive.zip", "w", zipfile.ZIP_DEFLATED) as zf:
    for path in Path("include_all_of_this_folder").rglob("*"):
        zf.write(path, path.as_posix())
回复 原文
爱的故事 2024-08-20 09:22:34

这是一种使用 pathlib 和上下文管理器的现代方法。将文件直接放入 zip 中，而不是放入子文件夹中。

def zip_dir(filename: str, dir_to_zip: pathlib.Path):
    with zipfile.ZipFile(filename, 'w', zipfile.ZIP_DEFLATED) as zipf:
        # Use glob instead of iterdir(), to cover all subdirectories.
        for directory in dir_to_zip.glob('**'):
            for file in directory.iterdir():
                if not file.is_file():
                    continue
                # Strip the first component, so we don't create an uneeded subdirectory
                # containing everything.
                zip_path = pathlib.Path(*file.parts[1:])
                # Use a string, since zipfile doesn't support pathlib  directly.
                zipf.write(str(file), str(zip_path))

Here's a modern approach, using pathlib, and a context manager. Puts the files directly in the zip, rather than in a subfolder.

def zip_dir(filename: str, dir_to_zip: pathlib.Path):
    with zipfile.ZipFile(filename, 'w', zipfile.ZIP_DEFLATED) as zipf:
        # Use glob instead of iterdir(), to cover all subdirectories.
        for directory in dir_to_zip.glob('**'):
            for file in directory.iterdir():
                if not file.is_file():
                    continue
                # Strip the first component, so we don't create an uneeded subdirectory
                # containing everything.
                zip_path = pathlib.Path(*file.parts[1:])
                # Use a string, since zipfile doesn't support pathlib  directly.
                zipf.write(str(file), str(zip_path))
回复 原文
凉栀 2024-08-20 09:22:34

我通过将 Mark Byers 的解决方案与 Reimund 和 Morten Zilmer 的评论（相对路径并包括空目录）合并来准备一个函数。作为最佳实践，with 用于 ZipFile 的文件构造。

该函数还准备一个默认的 zip 文件名，其中包含压缩目录名称和“.zip”扩展名。因此，它仅适用于一个参数：要压缩的源目录。

import os
import zipfile

def zip_dir(path_dir, path_file_zip=''):
if not path_file_zip:
    path_file_zip = os.path.join(
        os.path.dirname(path_dir), os.path.basename(path_dir)+'.zip')
with zipfile.ZipFile(path_file_zip, 'wb', zipfile.ZIP_DEFLATED) as zip_file:
    for root, dirs, files in os.walk(path_dir):
        for file_or_dir in files + dirs:
            zip_file.write(
                os.path.join(root, file_or_dir),
                os.path.relpath(os.path.join(root, file_or_dir),
                                os.path.join(path_dir, os.path.pardir)))

I prepared a function by consolidating Mark Byers' solution with Reimund and Morten Zilmer's comments (relative path and including empty directories). As a best practice, with is used in ZipFile's file construction.

The function also prepares a default zip file name with the zipped directory name and '.zip' extension. Therefore, it works with only one argument: the source directory to be zipped.

import os
import zipfile

def zip_dir(path_dir, path_file_zip=''):
if not path_file_zip:
    path_file_zip = os.path.join(
        os.path.dirname(path_dir), os.path.basename(path_dir)+'.zip')
with zipfile.ZipFile(path_file_zip, 'wb', zipfile.ZIP_DEFLATED) as zip_file:
    for root, dirs, files in os.walk(path_dir):
        for file_or_dir in files + dirs:
            zip_file.write(
                os.path.join(root, file_or_dir),
                os.path.relpath(os.path.join(root, file_or_dir),
                                os.path.join(path_dir, os.path.pardir)))
回复 原文
眼前雾蒙蒙 2024-08-20 09:22:34
# import required python modules
# You have to install zipfile package using pip install

import os,zipfile

# Change the directory where you want your new zip file to be

os.chdir('Type your destination')

# Create a new zipfile ( I called it myfile )

zf = zipfile.ZipFile('myfile.zip','w')

# os.walk gives a directory tree. Access the files using a for loop

for dirnames,folders,files in os.walk('Type your directory'):
    zf.write('Type your Directory')
    for file in files:
        zf.write(os.path.join('Type your directory',file))

# import required python modules
# You have to install zipfile package using pip install

import os,zipfile

# Change the directory where you want your new zip file to be

os.chdir('Type your destination')

# Create a new zipfile ( I called it myfile )

zf = zipfile.ZipFile('myfile.zip','w')

# os.walk gives a directory tree. Access the files using a for loop

for dirnames,folders,files in os.walk('Type your directory'):
    zf.write('Type your Directory')
    for file in files:
        zf.write(os.path.join('Type your directory',file))
回复 原文
听风念你 2024-08-20 09:22:34

嗯，在阅读了这些建议后，我想出了一种非常类似的方法，可以在 2.7.x 上使用，而无需创建“有趣的”目录名称（绝对类似的名称），并且只会在 zip 中创建指定的文件夹。

或者以防万一您需要 zip 中包含一个包含所选目录内容的文件夹。

def zipDir( path, ziph ) :
 """
 Inserts directory (path) into zipfile instance (ziph)
 """
 for root, dirs, files in os.walk( path ) :
  for file in files :
   ziph.write( os.path.join( root, file ) , os.path.basename( os.path.normpath( path ) ) + "\\" + file )

def makeZip( pathToFolder ) :
 """
 Creates a zip file with the specified folder
 """
 zipf = zipfile.ZipFile( pathToFolder + 'file.zip', 'w', zipfile.ZIP_DEFLATED )
 zipDir( pathToFolder, zipf )
 zipf.close()
 print( "Zip file saved to: " + pathToFolder)

makeZip( "c:\\path\\to\\folder\\to\\insert\\into\\zipfile" )

Well, after reading the suggestions I came up with a very similar way that works with 2.7.x without creating "funny" directory names (absolute-like names), and will only create the specified folder inside the zip.

Or just in case you needed your zip to contain a folder inside with the contents of the selected directory.

def zipDir( path, ziph ) :
 """
 Inserts directory (path) into zipfile instance (ziph)
 """
 for root, dirs, files in os.walk( path ) :
  for file in files :
   ziph.write( os.path.join( root, file ) , os.path.basename( os.path.normpath( path ) ) + "\\" + file )

def makeZip( pathToFolder ) :
 """
 Creates a zip file with the specified folder
 """
 zipf = zipfile.ZipFile( pathToFolder + 'file.zip', 'w', zipfile.ZIP_DEFLATED )
 zipDir( pathToFolder, zipf )
 zipf.close()
 print( "Zip file saved to: " + pathToFolder)

makeZip( "c:\\path\\to\\folder\\to\\insert\\into\\zipfile" )
回复 原文
彻夜缠绵 2024-08-20 09:22:34

创建 zip 文件的函数。

def CREATEZIPFILE(zipname, path):
    #function to create a zip file
    #Parameters: zipname - name of the zip file; path - name of folder/file to be put in zip file

    zipf = zipfile.ZipFile(zipname, 'w', zipfile.ZIP_DEFLATED)
    zipf.setpassword(b"password") #if you want to set password to zipfile

    #checks if the path is file or directory
    if os.path.isdir(path):
        for files in os.listdir(path):
            zipf.write(os.path.join(path, files), files)

    elif os.path.isfile(path):
        zipf.write(os.path.join(path), path)
    zipf.close()

Function to create zip file.

def CREATEZIPFILE(zipname, path):
    #function to create a zip file
    #Parameters: zipname - name of the zip file; path - name of folder/file to be put in zip file

    zipf = zipfile.ZipFile(zipname, 'w', zipfile.ZIP_DEFLATED)
    zipf.setpassword(b"password") #if you want to set password to zipfile

    #checks if the path is file or directory
    if os.path.isdir(path):
        for files in os.listdir(path):
            zipf.write(os.path.join(path, files), files)

    elif os.path.isfile(path):
        zipf.write(os.path.join(path), path)
    zipf.close()
回复 原文
罪#恶を代价 2024-08-20 09:22:34

总结其他，做一个函数：

import zipfile

def zipFolder(toZipFolder, outputZipFile):
  """
    zip/compress a whole folder/directory to zip file
  """
  print("Zip for foler %s" % toZipFolder)
  with zipfile.ZipFile(outputZipFile, 'w', zipfile.ZIP_DEFLATED) as zipFp:
    for dirpath, dirnames, filenames in os.walk(toZipFolder):
      # print("%s" % ("-"*80))
      # print("dirpath=%s, dirnames=%s, filenames=%s" % (dirpath, dirnames, filenames))
      # print("Folder: %s, Files: %s" % (dirpath, filenames))
      for curFileName in filenames:
        # print("curFileName=%s" % curFileName)
        curFilePath = os.path.join(dirpath, curFileName)
        # print("curFilePath=%s" % curFilePath)
        fileRelativePath = os.path.relpath(curFilePath, toZipFolder)
        # print("fileRelativePath=%s" % fileRelativePath)
        # print("  %s" % fileRelativePath)
        zipFp.write(curFilePath, arcname=fileRelativePath)
  print("Completed zip file %s" % outputZipFile)

调用：

  toZipFullPath = "/Users/crifan/dev/dev_root/iosReverse/WhatsApp/ipa/forRepackIpa_20231128"
  outputZipFile = "/Users/crifan/dev/dev_root/iosReverse/WhatsApp/ipa/WhatsApp_v23.24.0_20231128.ipa"
  zipFolder(toZipFullPath, outputZipFile)

输出：

Zip for foler /Users/crifan/dev/dev_root/iosReverse/WhatsApp/ipa/forRepackIpa_20231128
Completed zip file /Users/crifan/dev/dev_root/iosReverse/WhatsApp/ipa/WhatsApp_v23.20.79_idaAllSymbols_20231128.ipa

最新代码： crifanLib/crifanFile.py

summary others, make a function:

import zipfile

def zipFolder(toZipFolder, outputZipFile):
  """
    zip/compress a whole folder/directory to zip file
  """
  print("Zip for foler %s" % toZipFolder)
  with zipfile.ZipFile(outputZipFile, 'w', zipfile.ZIP_DEFLATED) as zipFp:
    for dirpath, dirnames, filenames in os.walk(toZipFolder):
      # print("%s" % ("-"*80))
      # print("dirpath=%s, dirnames=%s, filenames=%s" % (dirpath, dirnames, filenames))
      # print("Folder: %s, Files: %s" % (dirpath, filenames))
      for curFileName in filenames:
        # print("curFileName=%s" % curFileName)
        curFilePath = os.path.join(dirpath, curFileName)
        # print("curFilePath=%s" % curFilePath)
        fileRelativePath = os.path.relpath(curFilePath, toZipFolder)
        # print("fileRelativePath=%s" % fileRelativePath)
        # print("  %s" % fileRelativePath)
        zipFp.write(curFilePath, arcname=fileRelativePath)
  print("Completed zip file %s" % outputZipFile)

call:

  toZipFullPath = "/Users/crifan/dev/dev_root/iosReverse/WhatsApp/ipa/forRepackIpa_20231128"
  outputZipFile = "/Users/crifan/dev/dev_root/iosReverse/WhatsApp/ipa/WhatsApp_v23.24.0_20231128.ipa"
  zipFolder(toZipFullPath, outputZipFile)

output:

Zip for foler /Users/crifan/dev/dev_root/iosReverse/WhatsApp/ipa/forRepackIpa_20231128
Completed zip file /Users/crifan/dev/dev_root/iosReverse/WhatsApp/ipa/WhatsApp_v23.20.79_idaAllSymbols_20231128.ipa

latest code: crifanLib/crifanFile.py

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