从 appengine 数据存储区压缩一组 blob

发布于 2024-11-17 19:58:56 字数 1132 浏览 2 评论 0原文

我需要压缩数据存储中可用的一组 blob。这些可以是不同类型的，例如一些 html/images/swf/ 等。其中所有这些都可以作为 blob 在数据存储中使用。

我尝试实施这个解决方案：在 App Engine (Python) 中压缩动态文件？

尝试使用一些静态文本效果很好，我还可以使用一组具有各自内容的文件创建一个 zip，但在从查询生成 zip 时我无法找出一些问题。

z.writestr(fil.Template_name, my_data.encode('UTF-8'))
File "C:\Python25\lib\zipfile.py", line 626, in writestr
self.fp.write(zinfo.FileHeader())
File "C:\Python25\lib\zipfile.py", line 260, in FileHeader
return header + self.filename + extra
UnicodeDecodeError: 'ascii' codec can't decode byte 0xde in position 12: ordinal not in range(128)

这是这部分代码的错误

       class filesDB(db.model) 
                   Template_file = db.BlobProperty()
                   Template_name= db.StringProperty()


       output = StringIO.StringIO()
       z = zipfile.ZipFile(output,'w')
       files =  filesDB.all().filter("fCreatedBy","sandeep")
       for fil in files:
        my_data = fil.Template_file
        z.writestr(fil.Template_name, my_data)
        z.close()

原文

I need to zip a set of blobs available in data store.
These can of different types like some html/images/swf/ etc.
where all these are available in datastore as blob.

I tried to implement this solution:
Zipping dynamic files in App Engine (Python)?

Tried with some static texts it worked great, I am also able to create a zip with a set of files with respective content but I could not trace out some issue when making zip from query.

z.writestr(fil.Template_name, my_data.encode('UTF-8'))
File "C:\Python25\lib\zipfile.py", line 626, in writestr
self.fp.write(zinfo.FileHeader())
File "C:\Python25\lib\zipfile.py", line 260, in FileHeader
return header + self.filename + extra
UnicodeDecodeError: 'ascii' codec can't decode byte 0xde in position 12: ordinal not in range(128)

This is the error for this part of code

       class filesDB(db.model) 
                   Template_file = db.BlobProperty()
                   Template_name= db.StringProperty()


       output = StringIO.StringIO()
       z = zipfile.ZipFile(output,'w')
       files =  filesDB.all().filter("fCreatedBy","sandeep")
       for fil in files:
        my_data = fil.Template_file
        z.writestr(fil.Template_name, my_data)
        z.close()

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绝影如岚 2024-11-24 19:58:56

根据 zipfile 文档：

ZIP 文件没有正式的文件名编码。
如果您有 unicode 文件名，则必须将它们转换为所需编码的字节字符串，然后再将它们传递给 write()。

尝试将文件名编码为 UTF-8，例如：

class filesDB(db.model) 
                   Template_file = db.BlobProperty()
                   Template_name= db.StringProperty()


       output = StringIO.StringIO()
       z = zipfile.ZipFile(output,'w')
       files =  filesDB.all().filter("fCreatedBy","sandeep")
       for fil in files:
        my_data = fil.Template_file
        z.writestr(fil.Template_name.encode('utf-8'), my_data)
        z.close()

Per the zipfile documentation:

There is no official file name encoding for ZIP files.
If you have unicode file names, you must convert them to byte strings in your desired encoding before passing them to write().

Try to encode your filename in UTF-8 for example with:

class filesDB(db.model) 
                   Template_file = db.BlobProperty()
                   Template_name= db.StringProperty()


       output = StringIO.StringIO()
       z = zipfile.ZipFile(output,'w')
       files =  filesDB.all().filter("fCreatedBy","sandeep")
       for fil in files:
        my_data = fil.Template_file
        z.writestr(fil.Template_name.encode('utf-8'), my_data)
        z.close()

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