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这很奇怪，因为当我将一个部分的相同代码复制到另一个部分时，我不知道为什么会返回此错误，只更改一些通用视图（UpdateView）的标题，并且从不执​​行 get 函数以返回我的表单进行编辑一个客户，这就是代码：

网址

from django.conf.urls import patterns, include, url
from django.contrib.staticfiles.urls import staticfiles_urlpatterns
from . import views
from . import forms
from . import invoice


urlpatterns = patterns('',
    url(r'^pedidos/',views.PedidosListView.as_view(),name="pedidos_list"),
    url(r'^pedidos/add',views.add_pedido, name="pedido_add"),
    url(r'^pedidos/edit/(?P<pedido_id>\d+)',views.edit_pedido, name="pedido_edit"),
    url(r'^pedidos/delete/(?P<pedido_id>\d+)',views.delete_pedido, name="pedido_delete"),
    url(r'^pedido/(?P<pk>\d+)',forms.DetailPedido.as_view(), name="pedido_detail"),
    url(r'^pedido-pdf/(?P<pk>\d+)',invoice.detalle_pedido_pdf, name="pedido_detail_pdf"),
    url(r'^clientes/',views.ClientesListView.as_view(),name="clientes_list"),
    url(r'^clientes/edit/(?P<pk>\d+)$',forms.ClienteUpdateView.as_view(), name="clientes_edit"),
    url(r'^empleados/edit/(?P<pk>\d+)$',forms.EmpleadoUpdateView.as_view(),name="edit"),
    url(r'^empleados/',views.EmpleadoListView.as_view(),name="list"),

)

形式

class ClienteUpdateView(UpdateView):
    form_class = ClienteModelForm
    model = Cliente
    template_name = 'ventas/form.html'

    def get(self, request, **kwargs):
        self.object = Cliente.objects.get(pk=self.kwargs['pk'])
        form_class = self.get_form_class()
        form = self.get_form(form_class)
        context = self.get_context_data(object=self.object, form=form)
        return self.render_to_response(context)

    def form_valid(self, form):
        self.object = form.save(commit=False)
        self.object.save()
        return HttpResponseRedirect('ventas/clientes_list.html')

    def form_invalid(self,form):
      if self.request.is_ajax():
          return HttpResponseBadRequest(json.dumps(form.errors),
                                       mimetype="application/json")

javascript

$(".edit").click(function(ev){
    ev.preventDefault();
    var url = $(this).data('form');
    $.ajax({
        url: url,
        success: function(data, status){
            $('body').append($('<div id="myModal" class="modal hide fade">' + data + '</div>').modal());
            $("#myModal").find(".datePicker" ).datepicker({ dateFormat: "dd/mm/yy" });
        },
        error: function (xhr, ajaxOptions, thrownError) {
            console.log(xhr.status);
            console.log(xhr.responseText);
            console.log(thrownError);
        }
    })

    return false; // prevent the click propagation
});

html

  <button data-form="{% url 'ventas:clientes_edit' c.id %}" class="edit btn btn-mini btn-info">

如果在 ajax url 中更改 'ventas/empleados/edit/someid' 的值（与我为 do 复制的相同ClienteUpdateView，我可以检索员工编辑的表单，但我不能为客户做同样的事情，我错过了一些东西我现在没注意到？这让我发疯了！有什么想法吗？

问候！

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0 回答 0