如果我撰寫以下代碼：
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 Codepath用紅線標記了第二個函式。
這里有什么問題？
uj5u.com熱心網友回復：
這是您的代碼：-
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')
]
你的問題是你需要添加 (coma)" , " 畢竟路徑功能。A 并且您需要將您的最后一個路徑 "about/" 更改為 "contact/" 。
這是對的：-
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')
]
uj5u.com熱心網友回復：
你寫的代碼不正確：你正在處理一個串列，你應該在不同的路徑之間使用逗號，所以：
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')
]其他問題是您沒有在末尾使用斜杠（這不是強制性的，但建議使用）。最后，您的兩條路徑具有相同的名稱，因此可能about應該是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')
]轉載請註明出處，本文鏈接：https://www.uj5u.com/qita/360194.html