好吧,我几乎什么都试过了,但我不能让它工作。

我有一个带有ImageField的Django模型 我有通过HTTP下载图像的代码(测试和工作) 图像直接保存到'upload_to'文件夹中(upload_to是在ImageField中设置的) 我所需要做的就是将已经存在的图像文件路径与ImageField关联起来

我用六种不同的方式写了这段代码。

The problem I'm running into is all of the code that I'm writing results in the following behavior: (1) Django will make a 2nd file, (2) rename the new file, adding an _ to the end of the file name, then (3) not transfer any of the data over leaving it basically an empty re-named file. What's left in the 'upload_to' path is 2 files, one that is the actual image, and one that is the name of the image,but is empty, and of course the ImageField path is set to the empty file that Django try to create.

如果你不清楚,我将尝试说明:

## Image generation code runs.... 
/Upload
     generated_image.jpg     4kb

## Attempt to set the ImageField path...
/Upload
     generated_image.jpg     4kb
     generated_image_.jpg    0kb

ImageField.Path = /Upload/generated_image_.jpg

我怎样才能做到这一点而不让Django重新存储文件呢?我真正想要的是这样的东西……

model.ImageField.path = generated_image_path

...当然这是行不通的。

是的,我已经看了这里的其他问题,比如这个问题,以及django doc on File

更新 经过进一步的测试,它只有在Windows Server上的Apache下运行时才会执行此行为。当在XP的“runserver”下运行时,它不会执行此行为。

我被难住了。

下面是在XP上成功运行的代码…

f = open(thumb_path, 'r')
model.thumbnail = File(f)
model.save()

当前回答

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()

其他回答

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()

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。

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

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')

这可能不是你想要的答案。但是您可以使用charfield来存储文件的路径,而不是ImageFile。通过这种方式,您可以通过编程方式将上传的图像关联到字段,而无需重新创建文件。