Question

我的网址标记一直出错。语法是正确的，我尝试了不同的解决方案，在SO上问但没有运气。

my urls.py

urlpatterns = [
    url(r'^$', LandingView.as_view(), name="landing_page"),
    url(r'^add/$', AddView.as_view(), name="add"),
    url(r'^rent/(?P<slug>\w+)/$', rent_detail, name="rent_detail"),
    url(r'^add/space/$', AddSpaceView.as_view(), name="addSpace"),
    url(r'^lang/$', Language.as_view(), name="lang"),
    url(r'^upload/image/(?P<pk>\d+)/$', ImageUpload, name="ImageUpload"),
    url(r'^filter/space/$', FilterSpace.as_view(), name="filterSpace"),
    url(r'^api/', include(v1_api.urls)),
]

views.py

def ImageUpload(self, request, *args, **kwargs):
    try:
        rental = Rental.objects.get(pk = kwargs.get('pk'))
    except Rental.DoesNotExist:
        error_dict = {'message': 'Rental spae not found'}
        return self.render(request,'rentals/add.html',error_dict)
    for file in request.FILES.getlist('image'):
        image = GalleryImage.objects.create(image=file,rental=rental)
    response_dict = {
        'message': 'File Uploaded Successfully'
    }
    context = {
        'rental':rental
    }
    return self.render(request,'rentals/add.html',context)

class AddView(TemplateView):
    template_name = 'rentals/add.html'

add.html

<div id="listing">
</div>
<script type="text/javascript">
  // want to passs url for uploading image in ajax url parameter
    var data = {
        urltag: "{% url 'ImageUpload' rental.id %}"
    }

    console.log('url is', data); 
    $(function() {
      app.showListingSpaceForm("listing",data);
    });

</script>

用于上传图片的ajax

 $.ajax({
        url:"/upload/image/", // upload form is in add.html which is developed using frontend library reactjs and i need to pass id here so that the image will be saved to its associated rent.
        data:image,
        contentType:false,
        processData:false,
        type:'POST',
        mimeType: "multipart/form-data",
        success: function(data) {
          console.log('success');
        }
       });
    }

更新：Ajax代码，我希望在网址中使用ID将图片上传到相关的租金

Answer 1

当你转到/add/时，Django会运行AddView，因为你有一个网址模式。

url(r'^add/$', AddView.as_view(), name="add"),

该视图将呈现rentals/add.html，因为您有：

class AddView(TemplateView):
    template_name = 'rentals/add.html'

渲染模板会失败，因为您在网址标记中使用了rental，但您尚未在任何地方定义租借。

{% url 'ImageUpload' rental.id %}

除非您的AddView可以在模板上下文中设置rental，否则使用这样的网址标记将无效。

Answer 2

您需要明确传递kwarg参数，因为您在urls.py中定义了它：

var data = {
    urltag: "{% url 'ImageUpload' pk=rental.id %}"
}

NoReverseMatch at / add /

2 个答案: