Django:如何使用ImageField的upload_to属性

Question

我正在编写一个带图像上传的大型应用程序.

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()
    # ...

以下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()

这段代码违反了DRY原则 - 路径gallery重复两次.

问题:如何重用我在model(upload_to='gallery')中指定的路径,以便我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!

解决方案基于Paulo的答案

保存模型实例时,文件会自动上传,所以我只需要这样做:

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

该表格框架应该照顾这对你.除非您想将它们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})