在django中动态设置上传路径

参考于:http://scottbarnham.com/blog/2008/08/25/dynamic-upload-paths-in-django/

FileField或ImageField的参数upload_to可以是一个可执行的对象。这个可执行的函数可传入当前model实例与上传文件名,且必须返回一个路径。

下面是一个实例:

 

import os
from django.db import models

def get_image_path(instance, filename):
    return os.path.join('photos', str(instance.id), filename)

class Photo(models.Model):
    image = models.ImageField(upload_to=get_image_path)
 

get_image_path是一个可执行的对象(实际上是函数)。它从Photo的实例中获取id,并将它用于上传路径中。图像上传路径如:photos/1/kitty.jpg

可以使用实例中的任何字段或相关models中的字段。

为了使用get_image_path可用于多个models,可以作如下改进:

 

def get_image_path(path, attribute):
    def upload_callback(instance, filename):
        return '%s%s/%s' % (path, unicode(slugify(getattr(instance, attribute))), filename)
    return upload_callback

...
#Model definitions
class Photo(models.Model):
    name = models.CharField()
    image = models.ImageField(upload_to = get_image_path('uploads/', 'name'))

class Screenshot(models.Model):
    title = models.CharField()
    image = models.ImageField(upload_to = get_image_path('uploads/', 'title'))
 

 

为了更灵活,还可以作如下改进:

 

def get_image_path(path, attribute):
    def upload_callback(instance, filename):
        attributes = attribute.split('.')
        for attr in attributes:
            try:
                instance = getattr(instance,attr)
            except:
                print 'attribute error'
        return '%s%s/%s' % (path,unicode(instance),filename)
    return upload_callback

class Block(models.Model):
    description = models.CharField(max_length=255) 
class Building(models.Model):
    block = models.ForeignKey(Block)
    address = models.CharField(max_length=255)
    image = models.ImageField(upload_to=get_image_path('uploads/','block.description'))
 

作了如此改进后,可以方便的设置与当前实例相关的models中字段了。

 

注意的是如果你使用id,得先确定model实例在上传文件之前保存了,否则id会不存在。为了解决这个情况,在view中可以如下操作

...
    if request.method == 'POST':
        form = PhotoForm(request.POST, request.FILES)
        if form.is_valid():
            photo = Photo.objects.create()
            image_file = request.FILES['image']
            photo.image.save(image_file.name, image_file)
...
 

 

你可能感兴趣的:(django,OS,Blog)