我正在尝试将 UploadedFile 转换为 PIL Image 对象以对其进行缩略图，然后将我的缩略图函数返回的 PIL Image 对象转换回 File 对象。我该怎么做？
【问题讨论】：
Image 实例到一个 Django File 实例。 Django 的File 是Python 的File 类的子类。
标签： python django python-imaging-library django-file-upload django-uploads
我必须分几步完成，php 中的 imagejpeg() 需要类似的过程。并不是说没有办法将内容保存在内存中，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）。
型号：
class YourModel(Model):
img = models.ImageField(upload_to='photos')
thumb = models.ImageField(upload_to='thumbs')
用法：
#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)
【讨论】：
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。这是一个关于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。我现在在我的项目中工作，使用 django-storages 上传到 S3。这花了我一天的大部分时间在此处正确找到解决方案。
【讨论】：
StringIO 在 Python 3 中工作的？我不得不使用BytesIO（并且由于它没有 len() 函数，因此必须使用 tell() 来确定它的长度。
BytesIO 在这里尝试我的答案stackoverflow.com/a/30435175/3033586 顺便说一句，之前的评论已过时并被删除
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}))
])
如果您想保留旧文件，请确保将keep_orphans=True 传递给该字段，否则它们会在替换时被清理。
【讨论】：
对于那些使用django-storages/-redux 将图像文件存储在 S3 上的用户，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()
【讨论】：
这是 python 3.5 和 django 1.10
的实际工作示例在views.py中：
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(...)
【讨论】：
为 Python 3+ 整合 cmets 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())
【讨论】：
为那些像我一样想要将它与 Django 的FileSystemStorage 结合起来的人完成：
（我在这里做的是上传一张图片，将其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()
【讨论】：