Question

我正在编写一个包含图片上传的大型应用程序。

这是我的模特：

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()

此代码违反了DRY原则 - 路径gallery重复两次。

问题：如何重用我在模型中指定的路径（upload_to='gallery'），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！

基于Paulo回答的解决方案

当保存模型实例时，文件会自动上传，所以我只需这样做：

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

forms framework应该为您解决此问题。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})

Django：如何使用ImageField的upload_to属性

1 个答案: