python-3.x - 多参数 url 模式 django 2.0

Question

我想在我的 url 模式中传递两个参数，但我收到错误无反向匹配，即“项目”。虽然它只有一个参数可以正常工作。

这是主要的网址文件-

urlpatterns = [
    path('admin/', admin.site.urls),
    path(r'^materials/(?P<name>(\s+)/',include('materials.urls')),
    path(r'^projects/',include('projects.urls')),
]

项目.urls-

urlpatterns = [
path('',views.view_projects,name='view_projects'),
path('(?<projectid>\d+)/',views.project_steps,name='project_steps'),
path('(P<projectid>\d+)/(P<stepid>\d+)/',views.project_steps,
name='project_steps'),
] 

View .py-

 def view_projects(request):
   projects = project.objects.all
   return render(request,'projects/project_view.html', 
   {'projects':projects})

def project_steps(request,projectid,stepno=1):
  projects = project.objects.all
  stepss = steps.objects.all
  return render(request,'projects/project_steps.html', 
  {'projectid':projectid,'steps':stepss,'projects':projects,
  'stepno':stepno})

模板-

 "{% url 'projects' projectid=project.id stepno=step.step_no %}"

Answer 1

您可以执行以下操作。

Old Way
(r'^view_url/(\d+)/(\d+)$', r'app_name.views.view_function'),
def view_function(request, param1, param2):
    """
    :param request:
    :param param1:
    :param param2:
    :return:
    """
    return render('/* template path and parameters */')

New Way
(r'^view_url/<int:param1>/<int:param2>$', r'app_name.views.view_function'),
def view_function(request, param1, param2):
    """
    :param request:
    :param param1:
    :param param2:
    :return:
    """
    return render('/* template path and parameters */')

有关 django 2.0 中正则表达式模式的更多详细信息，您可以查看 django 文档链接。
https://docs.djangoproject.com/en/2.1/topics/http/urls/