下面是一个工作良好的方法，允许您将文件转换为某种格式(以避免“不能将模式P写入JPEG”错误):

import urllib2
from django.core.files.base import ContentFile
from PIL import Image
from StringIO import StringIO

def download_image(name, image, url):
    input_file = StringIO(urllib2.urlopen(url).read())
    output_file = StringIO()
    img = Image.open(input_file)
    if img.mode != "RGB":
        img = img.convert("RGB")
    img.save(output_file, "JPEG")
    image.save(name+".jpg", ContentFile(output_file.getvalue()), save=False)

这里的image是django的ImageField还是your_model_instance.image 下面是一个用法示例:

p = ProfilePhoto(user=user)
download_image(str(user.id), p.image, image_url)
p.save()

希望这能有所帮助

你可以使用Django REST框架和python请求库以编程方式将图像保存到Django ImageField中

下面是一个例子:

import requests


def upload_image():
    # PATH TO DJANGO REST API
    url = "http://127.0.0.1:8080/api/gallery/"

    # MODEL FIELDS DATA
    data = {'first_name': "Rajiv", 'last_name': "Sharma"}

    #  UPLOAD FILES THROUGH REST API
    photo = open('/path/to/photo', 'rb')
    resume = open('/path/to/resume', 'rb')
    files = {'photo': photo, 'resume': resume}

    request = requests.post(url, data=data, files=files)
    print(request.status_code, request.reason) 

如果模型还没有创建，超级简单:

首先，将图像文件复制到上传路径(在下面的代码片段中假设= 'path/')。

其次，使用如下语句:

class Layout(models.Model):
    image = models.ImageField('img', upload_to='path/')

layout = Layout()
layout.image = "path/image.png"
layout.save()

在django 1.4中测试和工作，它可能也适用于现有的模型。

我有一些代码，从网络上获取图像，并将其存储在模型中。重要的部分是:

from django.core.files import File  # you need this somewhere
import urllib


# The following actually resides in a method of my model

result = urllib.urlretrieve(image_url) # image_url is a URL to an image

# self.photo is the ImageField
self.photo.save(
    os.path.basename(self.url),
    File(open(result[0], 'rb'))
    )

self.save()

这有点令人困惑，因为它是从我的模型中提取出来的，有点脱离上下文，但重要的部分是:

从web中提取的图像不存储在upload_to文件夹中，而是由urllib.urlretrieve()存储为tempfile，然后丢弃。 ImageField.save()方法接受一个文件名(os.path.basename位)和一个django.core.files.File对象。

如果您有问题或需要澄清，请告诉我。

编辑:为了清晰起见，下面是模型(减去任何必需的import语句):

class CachedImage(models.Model):
    url = models.CharField(max_length=255, unique=True)
    photo = models.ImageField(upload_to=photo_path, blank=True)

    def cache(self):
        """Store image locally if we have a URL"""

        if self.url and not self.photo:
            result = urllib.urlretrieve(self.url)
            self.photo.save(
                    os.path.basename(self.url),
                    File(open(result[0], 'rb'))
                    )
            self.save()

class tweet_photos(models.Model):
upload_path='absolute path'
image=models.ImageField(upload_to=upload_path)
image_url = models.URLField(null=True, blank=True)
def save(self, *args, **kwargs):
    if self.image_url:
        import urllib, os
        from urlparse import urlparse
        file_save_dir = self.upload_path
        filename = urlparse(self.image_url).path.split('/')[-1]
        urllib.urlretrieve(self.image_url, os.path.join(file_save_dir, filename))
        self.image = os.path.join(file_save_dir, filename)
        self.image_url = ''
    super(tweet_photos, self).save()

我用uuid在django 2 python 3中保存图像，因为这是django如何做的:

import uuid   
from django.core.files import File 
import urllib

httpUrl = "https://miimgeurl/image.jpg"
result = urllib.request.urlretrieve(httpUrl)            
mymodel.imagefield.save(os.path.basename(str(uuid.uuid4())+".jpg"),File(open(result[0], 'rb')))
mymodel.save()