你如何将PIL`Image`转换为Django`File`？

Question

我正在尝试将UploadedFile一个PIL Image对象转换为缩略图,然后将Image我的缩略图函数返回的PIL 对象转换回一个File对象.我怎样才能做到这一点？

Answer 1

不必回写文件系统,然后通过公开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.
...

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Answer 2

我必须在几个步骤中执行此操作,php中的imagejpeg()需要类似的过程.不是说没有办法把东西保存在内存中,但是这个方法为你提供了原始图像和拇指的文件引用(如果你必须返回并改变你的拇指大小,通常是一个好主意).

1. 保存文件
2. 用PIL从文件系统打开它,
3. 使用PIL保存到临时目录,
4. 然后打开一个Django文件,让它工作.

模型:

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)

Answer 3

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(...)