就说一点。答案是可行的，但是，如果你在windows上工作，你可能想用'rb'打开()文件。是这样的:

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()

否则文件将在第一个0x1A字节处被截断。

你可以试试:

model.ImageField.path = os.path.join('/Upload', generated_image_path)

在Django 3中， 用这样一个模型:

class Item(models.Model):
   name = models.CharField(max_length=255, unique=True)
   photo= models.ImageField(upload_to='image_folder/', blank=True)

如果图片已经上传，我们可以直接做:

Item.objects.filter(...).update(photo='image_folder/sample_photo.png')

or

my_item = Item.objects.get(id=5)
my_item.photo='image_folder/sample_photo.png'
my_item.save()

我所做的是创建我自己的存储，只是不会将文件保存到磁盘:

from django.core.files.storage import FileSystemStorage

class CustomStorage(FileSystemStorage):

    def _open(self, name, mode='rb'):
        return File(open(self.path(name), mode))

    def _save(self, name, content):
        # here, you should implement how the file is to be saved
        # like on other machines or something, and return the name of the file.
        # In our case, we just return the name, and disable any kind of save
        return name

    def get_available_name(self, name):
        return name

然后，在我的模型中，对于我的ImageField，我使用了新的自定义存储:

from custom_storage import CustomStorage

custom_store = CustomStorage()

class Image(models.Model):
    thumb = models.ImageField(storage=custom_store, upload_to='/some/path')

工作! 您可以使用FileSystemStorage保存映像。 检查下面的示例

def upload_pic(request):
if request.method == 'POST' and request.FILES['photo']:
    photo = request.FILES['photo']
    name = request.FILES['photo'].name
    fs = FileSystemStorage()
##### you can update file saving location too by adding line below #####
    fs.base_location = fs.base_location+'/company_coverphotos'
##################
    filename = fs.save(name, photo)
    uploaded_file_url = fs.url(filename)+'/company_coverphotos'
    Profile.objects.filter(user=request.user).update(photo=photo)

如果您只想“设置”实际的文件名，而不需要加载和重新保存文件(!!)，或者使用charfield(!!)，您可能想尝试这样的操作——

model_instance.myfile = model_instance.myfile.field.attr_class(model_instance, model_instance.myfile.field, 'my-filename.jpg')

这将点亮model_instance.myfile。Url和所有其他的，就像你上传了文件一样。

就像@t-stone说的，我们真正想要的是能够设置instance.myfile.path = 'my-filename.jpg'，但是Django目前不支持这个功能。