python - Django:如何使用 ImageField 的 upload_to 属性

Question

我正在编写一个带有图片上传功能的大型应用程序。

这是我的模型:

class GallryImage(models.Model):
    # ...
    image   = models.ImageField(max_length=255, upload_to='gallery', height_field='width', width_field='height')
    width   = models.IntegerField()
    height  = models.IntegerField()
    # ...

这是我处理上传的方式:

image_name = 'image.png';
destination = open(settings.MEDIA_ROOT + '/gallery/' + image_name, 'wb+')
for chunk in f.chunks():
    destination.write(chunk)
destination.close()

这段代码有点违反了 DRY 原则——路径 gallery 重复了两次。

问题:如何重用我在模型中指定的路径 (upload_to='gallery')，这样我就不必在上传处理程序中重复？

我正在使用 python 2.6 和 Django 1.3 beta。

谢谢!

基于 Paulo 回答的解决方案

保存模型实例时，文件会自动上传，所以我要做的就是:

def add(request):
    from forms import ImageAddForm
    form = ImageAddForm()
    if request.method == 'POST':
        form = ImageAddForm(request.POST, request.FILES)
        if form.is_valid():
            image = GalleryImage(
                image   = form.cleaned_data['image']
            )
            image.save() # file is uploaded to upload_to dir!
            return HttpResponseRedirect(reverse('image_add') + '?image_added=')
    else:
        form = ImageAddForm()

    return render_to_response('gallery/add.html',
                              locals(),
                              context_instance=RequestContext(request))

Answer 1

forms framework应该为你处理这个。除非您想将文件存储在文件系统以外的某个容器中，否则无需手动保存文件。

class UploadImageForm(forms.ModelForm):
    class Meta:
        model = GallryImage
...
# Sample view
def upload_file(request):
    if request.method == 'POST':
        form = UploadImageForm(request.POST, request.FILES)
        if form.is_valid():
            form.save()
            return HttpResponseRedirect('/success/url/')
    else:
        form = UploadImageForm()
    return render_to_response('upload.html', {'form': form})