如何将 PIL `Image` 转换为 Django `File`？

Question

我正在尝试将 UploadedFile 转换为 PIL Image 对象以对其进行缩略图，然后转换我的缩略图函数返回的 PIL Image 对象返回到 File 对象。我该怎么做？

Answer 1

无需写回文件系统，然后通过 open 调用将文件带回内存的方法是使用 StringIO 和 Django InMemoryUploadedFile。以下是有关如何执行此操作的快速示例。这假设您已经有一个名为“thumb”的缩略图：

import StringIO

from django.core.files.uploadedfile import InMemoryUploadedFile

# Create a file-like object to write thumb data (thumb data previously created
# using PIL, and stored in variable 'thumb')
thumb_io = StringIO.StringIO()
thumb.save(thumb_io, format='JPEG')

# Create a new Django file-like object to be used in models as ImageField using
# InMemoryUploadedFile.  If you look at the source in Django, a
# SimpleUploadedFile is essentially instantiated similarly to what is shown here
thumb_file = InMemoryUploadedFile(thumb_io, None, 'foo.jpg', 'image/jpeg',
                                  thumb_io.len, None)

# Once you have a Django file-like object, you may assign it to your ImageField
# and save.
...

如果您需要更多说明，请告诉我。我现在正在我的项目中使用此功能，使用 django-storages 上传到 S3。我花了一天的时间才在这里正确找到解决方案。

Answer 2

我必须通过几个步骤来完成此操作，php 中的 imagejpeg() 需要类似的过程。并不是说没有办法将内容保留在内存中，但这种方法为您提供了对原始图像和拇指的文件引用（通常是一个好主意，以防您必须返回并更改拇指大小）。

1. 保存文件，
2. 使用 PIL 从文件系统打开它，
3. 使用 PIL 保存到临时目录，
4. 然后作为 Django 文件打开，这样才能正常工作。

型号：

class YourModel(Model):
    img = models.ImageField(upload_to='photos')
    thumb = models.ImageField(upload_to='thumbs')

用途：

#in upload code
uploaded = request.FILES['photo']
from django.core.files.base import ContentFile
file_content = ContentFile(uploaded.read())
new_file = YourModel() 
#1 - get it into the DB and file system so we know the real path
new_file.img.save(str(new_file.id) + '.jpg', file_content)
new_file.save()

from PIL import Image
import os.path

#2, open it from the location django stuck it
thumb = Image.open(new_file.img.path)
thumb.thumbnail(100, 100)

#make tmp filename based on id of the model
filename = str(new_file.id)

#3. save the thumbnail to a temp dir

temp_image = open(os.path.join('/tmp',filename), 'w')
thumb.save(temp_image, 'JPEG')

#4. read the temp file back into a File
from django.core.files import File
thumb_data = open(os.path.join('/tmp',filename), 'r')
thumb_file = File(thumb_data)

new_file.thumb.save(str(new_file.id) + '.jpg', thumb_file)

Answer 3

这是views.py中python 3.5和django 1.10的实际工作示例

：

from io import BytesIO
from django.core.files.base import ContentFile
from django.core.files.uploadedfile import InMemoryUploadedFile

def pill(image_io):
    im = Image.open(image_io)
    ltrb_border = (0, 0, 0, 10)
    im_with_border = ImageOps.expand(im, border=ltrb_border, fill='white')

    buffer = BytesIO()
    im_with_border.save(fp=buffer, format='JPEG')
    buff_val = buffer.getvalue()
    return ContentFile(buff_val)

def save_img(request)
    if request.POST:
       new_record = AddNewRecordForm(request.POST, request.FILES)
       pillow_image = pill(request.FILES['image'])
       image_file = InMemoryUploadedFile(pillow_image, None, 'foo.jpg', 'image/jpeg', pillow_image.tell, None)
       request.FILES['image'] = image_file  # really need rewrite img in POST for success form validation
       new_record.image = request.FILES['image']
       new_record.save()
       return redirect(...)

Answer 4

整理 Python 3+ 的评论和更新

from io import BytesIO
from django.core.files.base import ContentFile
import requests

# Read a file in

r = request.get(image_url)
image = r.content
scr = Image.open(BytesIO(image))

# Perform an image operation like resize:

width, height = scr.size
new_width = 320
new_height = int(new_width * height / width)
img = scr.resize((new_width, new_height))

# Get the Django file object

thumb_io = BytesIO()
img.save(thumb_io, format='JPEG')
photo_smaller = ContentFile(thumb_io.getvalue())

Answer 5

对于那些像我一样想要将其与 Django 的 FileSystemStorage 结合使用的人来说，需要完成以下内容：
（我在这里所做的是上传图像，将其大小调整为二维并保存两个文件。

utils.py

def resize_and_save(file):
    size = 1024, 1024
    thumbnail_size = 300, 300
    uploaded_file_url = getURLforFile(file, size, MEDIA_ROOT)
    uploaded_thumbnail_url = getURLforFile(file, thumbnail_size, THUMBNAIL_ROOT)
    return [uploaded_file_url, uploaded_thumbnail_url]

def getURLforFile(file, size, location):
    img = Image.open(file)
    img.thumbnail(size, Image.ANTIALIAS)
    thumb_io = BytesIO()
    img.save(thumb_io, format='JPEG')
    thumb_file = InMemoryUploadedFile(thumb_io, None, file.name, 'image/jpeg', thumb_io.tell, None)
    fs = FileSystemStorage(location=location)
    filename = fs.save(file.name, thumb_file)
    return fs.url(filename)  

在views.py中

if request.FILES:
        fl, thumbnail = resize_and_save(request.FILES['avatar'])
        #delete old profile picture before saving new one
        try:
            os.remove(BASE_DIR + user.userprofile.avatarURL)
        except Exception as e:
            pass         
        user.userprofile.avatarURL = fl
        user.userprofile.thumbnailURL = thumbnail
        user.userprofile.save()

Answer 6

这是一个可以做到这一点的应用程序： django-smartfields

from django.db import models

from smartfields import fields
from smartfields.dependencies import FileDependency
from smartfields.processors import ImageProcessor

class ImageModel(models.Model):
    image = fields.ImageField(dependencies=[
        FileDependency(processor=ImageProcessor(
            scale={'max_width': 150, 'max_height': 150}))
    ])

确保传递 keep_orphans=如果您想保留旧文件，则确实如此，否则它们会在替换时被清除。

Answer 7

对于那些使用 django-storages/-redux 在 S3 上存储图像文件的人，这是我采用的路径（下面的示例创建现有图像的缩略图）：

from PIL import Image
import StringIO
from django.core.files.storage import default_storage

try:
    # example 1: use a local file
    image = Image.open('my_image.jpg')
    # example 2: use a model's ImageField
    image = Image.open(my_model_instance.image_field)
    image.thumbnail((300, 200))
except IOError:
    pass  # handle exception

thumb_buffer = StringIO.StringIO()
image.save(thumb_buffer, format=image.format)
s3_thumb = default_storage.open('my_new_300x200_image.jpg', 'w')
s3_thumb.write(thumb_buffer.getvalue())
s3_thumb.close()