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

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()

其他回答

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) 
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。

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')