为什么在django框架vscode扩展pylance在path给我reed行

Question

如果我写以下代码：

from django.contrib import admin
from django.urls import path
from home import views
urlpatterns = [
path("/", views.home, name='home')
path("about", views.about, name='about')
path("services", views.services, name='services')
path("about", views.contact, name='contact')
]

VS Code用红线标记第二个path函数。

这里可能有什么问题？

Answer 1

这是您的代码：-

from django.contrib import admin
from django.urls import path
from home import views
urlpatterns = [
path("/", views.home, name='home')
path("about", views.about, name='about'),
path("services", views.services, name='services')
path("about", views.contact, name='contact')

]

你的问题是你需要添加(昏迷("毕竟路径函数。A.你需要改变你的最后一条路"；大约/"；至"；接触/"。

这是正确的：-

from django.contrib import admin
from django.urls import path
from home import views
urlpatterns = [
path("/", views.home, name='home')
path("about/", views.about, name='about'),
path("services/", views.services, name='services')
path("contact/", views.contact, name='contact')

]

Answer 2

您编写的代码不正确：您使用的是列表，您应该在不同的路径之间使用逗号，因此：

from django.contrib import admin
from django.urls import path
from home import views
urlpatterns = [
path("/", views.home, name='home'),  # ← comma
path("about", views.about, name='about'),  # ← comma
path("services", views.services, name='services'),  # ← comma
path("about", views.contact, name='contact')
]

其他问题是结尾处没有使用斜线(这不是强制性的，但建议使用(。最后，您的两个路径具有相同的名称，因此具有about的路径很可能应该是contact:

from django.contrib import admin
from django.urls import path
from home import views
urlpatterns = [
path("/", views.home, name='home'),
path("about/", views.about, name='about'),
path("services/", views.services, name='services'),
path("contact/", views.contact, name='contact')
]