当我输入127.0.0.1：8000/decore_detail/123 B/时，我想得到decore_detail.html。但是，我运行下面的配置代码，它向我报告错误。如何解决？“TyoeError：context必须是dict而不是set”
Traceback (most recent call last):
File "C:\Users\daiyij\Anaconda3\envs\django\lib\site-packages\django\core\handlers\exception.py", line 55, in inner
response = get_response(request)
File "C:\Users\daiyij\Anaconda3\envs\django\lib\site-packages\django\core\handlers\base.py", line 197, in _get_response
response = wrapped_callback(request, *callback_args, **callback_kwargs)
File "D:\decoredesigner\mySite\DecoreGenerator\views.py", line 14, in decore_detail
return render(request,'decore_detail.html',{'decore_detail',decore_detail})
File "C:\Users\daiyij\Anaconda3\envs\django\lib\site-packages\django\shortcuts.py", line 24, in render
content = loader.render_to_string(template_name, context, request, using=using)
File "C:\Users\daiyij\Anaconda3\envs\django\lib\site-packages\django\template\loader.py", line 62, in render_to_string
return template.render(context, request)
File "C:\Users\daiyij\Anaconda3\envs\django\lib\site-packages\django\template\backends\django.py", line 57, in render
context = make_context(
File "C:\Users\daiyij\Anaconda3\envs\django\lib\site-packages\django\template\context.py", line 278, in make_context
raise TypeError(
Exception Type: TypeError at /decore_detail/7520N/
Exception Value: context must be a dict rather than set.字符串
urls.py：
from myapp.views import decore_detail
urlpatterns = [
path('admin/', admin.site.urls),
path('decore_detail/<str:decore_detail>/',views.decore_detail,name='decore_detail'),
]型
views.py：
def decore_detail(request,decore_detail):
context = {"decore_detail": decore_detail}
return render(request,'decore_detail.html',{'decore_detail',decore_detail})型
decore_detail.py：
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8" />
<title>Item Detail</title>
</head>
<body>
<h1>Item Detail - {{decore_detail}}</h1>
</body>
</html>型
1条答案
按热度按时间fwzugrvs1#
你可能想做这样的事情：
网址：Views.py
字符串