Question

我正在开展一个项目，目前我的任务是允许用户上传新图片。我编辑了我在这里找到的教程中的代码：Need a minimal Django file upload example。我最初遵循教程，只是将“文件”更改为“图像”，因为我只允许图像上传，并且工作正常。然后，我试图将它折叠成现有的Django项目，我遇到了我的问题：虽然我已经包含了所有的错误块，但是当我尝试上传不是图像的东西时，我没有收到任何错误消息。我已经包含了将图像标记为“公共”的功能，但即使我将其取出，也不会显示错误消息。为了使这项工作与项目的其余部分一起工作，我不得不将这些内容分开，而不是在教程中，我想知道这是否是问题。

以下项目的相关代码。

模板

<form action="{% url "upload_image" %}" method="post" enctype="multipart/form-data">
    {% csrf_token %}
    <p>{{ form.non_field_errors }}</p>
    <p>{{ form.imagefile.label_tag }} {{ form.imagefile.help_text }}</p>
    <p>
        {{ form.imagefile.errors }}
        {{ form.imagefile }}
    </p>
    <p> {{ form.errors }}</p>
    <p>{{ form.public }} {{ form.public.label }}</p>
    <p><input type="submit" value="Upload" /></p>
</form>

表单

（我这里有两个表格，因为我无法弄清楚如何检查复选框的值。）

class ImageForm(Form):
    imagefile=ImageField(label="Select image to upload.")
    public=BooleanField(label="Public image", required=False)

class PublicImageForm(ImageForm):
    public=BooleanField(label="Public image", required=True)

查看以呈现包含表单的页面

def upload_image_view(request):
    if request.method=='GET':
        form=ImageForm()
        # TODO: Limits to images uploaded by current user
        my_images = UploadedImage.objects.filter(owner=User.objects.get(username=request.user))
        return render_to_response(
        'uploader.html',
        {'my_images': my_images, 'form': form,},
        context_instance=RequestContext(request))
    else:
        return HttpResponseNotAllowed(['GET'])

查看哪种表单

def upload_new_image(request):
    if request.method=='POST':
        form = ImageForm(request.POST, request.FILES)
        if form.is_valid():
            u= User.objects.get(username=request.user)

            form = PublicImageForm(request.POST, request.FILES)
            if form.is_valid():
                newdoc = UploadedImage(imagefile = request.FILES['imagefile'], owner=u, isPublic=True)
            else:
                newdoc = UploadedImage(imagefile = request.FILES['imagefile'], owner=u, )

            newdoc.save()
        # else:
            # form = ImageForm()
        return HttpResponseRedirect('/workspace/upload_image/')
    else:
        return HttpResponseNotAllowed(['POST'])

模型

class UploadedImage(models.Model):
    imagefile=models.ImageField(upload_to="user_images/uploaded_images")
    owner=models.ForeignKey(User)
    isPublic=models.BooleanField(default=False)

Answer 1

您只是在表单无效时重定向。所以任何错误都会丢失，什么都不会发生。我建议你把这两个观点结合起来。（没有理由将它们分开。）然后，当表单无效时，您可以返回相同的视图。这是我的粗略实现:(未经测试）

def upload_image_view(request):
    if request.method=='GET':
        form=ImageForm()
        # TODO: Limits to images uploaded by current user
        my_images = UploadedImage.objects.filter(owner=User.objects.get(username=request.user))
        return render_to_response(
            'uploader.html',
            {'my_images': my_images, 'form': form,},
            context_instance=RequestContext(request)
        )
    elif request.method=='POST':
        form = ImageForm(request.POST, request.FILES)
        if form.is_valid():
            u= User.objects.get(username=request.user)

            public_form = PublicImageForm(request.POST, request.FILES)
            if public_form.is_valid():
                newdoc = UploadedImage(imagefile = request.FILES['imagefile'], owner=u, isPublic=True)
            else:
                newdoc = UploadedImage(imagefile = request.FILES['imagefile'], owner=u, )

            newdoc.save()

            return HttpResponseRedirect('/workspace/upload_image/')
        else:
            my_images = UploadedImage.objects.filter(owner=User.objects.get(username=request.user))
            return render_to_response(
                'uploader.html',
                {'my_images': my_images, 'form': form,},
                context_instance=RequestContext(request)
            )
    else:
        return HttpResponseNotAllowed(['GET', 'POST'])

Django Image Upload表单不显示错误消息

1 个答案: