无法映射 Django URL

Question

I've completed all of the sections of the Django tutorial and have started my own project now to practice.我已经完成了 Django 教程的所有部分，现在已经开始了我自己的项目来练习。 I am back at the beginning tutorial where it talks about views/mapping urls .我回到开始教程，它讨论了 views/mapping urls 。 I also am following this tutorial for trying to display a table我也在关注本教程以尝试显示表格

For whatever reason, I cannot figure out why when I try to hit http://127.0.0.1:8000/show/ , it returns 404. I've been staring at this for the last hour and have been going back and forth between the tutorial and my code.无论出于何种原因，我无法弄清楚为什么当我尝试点击http://127.0.0.1:8000/show/ 时，它返回 404。我一直盯着这个看最后一个小时，并且一直在这之间来回走动教程和我的代码。 I had to do things a little bit differently than the 2nd mentioned tutorial, mainly that they didn't talk about creating an app-level urls.py file.我不得不做的事情与第二个提到的教程略有不同，主要是他们没有谈论创建应用程序级别的 urls.py 文件。 Everything up to this point has worked fine.到目前为止，一切都运行良好。 The models.py file created the table within the MySQL database, as I can see it in the workbench. models.py 文件在 MySQL 数据库中创建了表，正如我在工作台中看到的那样。

My project structure is like this:我的项目结构是这样的：

mywebsite (project)我的网站（项目）
displaydata (app)显示数据（应用程序）

Here is my project level urls.py file located in the mywebsite folder:这是位于 mywebsite 文件夹中的我的项目级 urls.py 文件：

from django.contrib import admin
from django.urls import include,path

urlpatterns = [
    path('admin/', admin.site.urls),
    path('displaydata/', include('displaydata.urls'))
]

Here is my app-level urls.py file located in the displaydata folder:这是位于 displaydata 文件夹中的我的应用程序级 urls.py 文件：

from django.urls import path
from . import views
app_name = 'displaydata'

urlpatterns = [
    path('', views.show, name='show')
]

Here is my displaydata views.py file:这是我的 displaydata views.py 文件：

from django.shortcuts import render,redirect
from django.http import HttpResponse
from .models import Shipment

# Create your views here.

def show(request):
    shipments = Shipment.objects.all()
    return HttpResponse(render(request,"show.html",{'shipment':shipments}))

Here is the show.html file:这是 show.html 文件：

<!DOCTYPE html>
<html lang="en">
<head>
  <title>Django CRUD Operations</title>
  <meta charset="utf-8">
  <link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.4.1/css/bootstrap.min.css">
  <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.4.1/jquery.min.js"></script>
  <script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.4.1/js/bootstrap.min.js"></script> 
</head>
<body>
<div class="container">
<table class="table table-striped">
    <thead>
      <tr>
        <th>Shipment ID</th>
        <th>Driver</th>
        <th>Destination City</th>
        <th>Destination State</th>
      </tr>
    </thead>
    <tbody>
    {% for ship in shipment %}  
      <tr>
        <td>{{ship.id}}</td>
        <td>{{ship.driver}}</td>
        <td>{{ship.destination_city}}</td>
        <td>{{ship.destination_state}}</td>
      </tr>
      {% endfor %} 
    </tbody>
</table>    
</div>
</body>
</html>

Answer 1

It will hit the show view for the URL 127.0.0.1:8000 /displaydata/ .它将点击 URL 127.0.0.1:8000 /displaydata/的show视图。

This the case because you include all the displaydata urls with the displaydata/ prefix.这是因为您包含所有带有displaydata/前缀的displaydata url。 In the url patterns of your displaydata app, there is one pattern: the empty string, so it will match this for the path /dispaydata .在您的displaydata应用程序的 url 模式中，有一个模式：空字符串，因此它将与路径/dispaydata 。

If you want to access the view with /show , you can use an empty string as prefix in the project urls:如果要使用/show访问视图，可以在项目 url 中使用空字符串作为前缀：

from django.contrib import admin
from django.urls import include,path

urlpatterns = [
    path('admin/', admin.site.urls),
    path('', include('displaydata.urls'))
]

and then for the displaydata urls work with:然后对于displaydata url 使用：

from django.urls import path
from . import views
app_name = 'displaydata'

urlpatterns = [
    path('/show/', views.show, name='show')
]

If the template is located at app_name /templates/ app_name /show.html , then you render the template with:如果模板位于app_name /templates/ app_name /show.html ，则使用以下命令渲染模板：

def show(request):
    shipments = Shipment.objects.all()
    return render(request,'app_name/show.html',{'shipment': shipments})

无法映射 Django URL

问题描述

1 个解决方案

解决方案1
1 已采纳 2021-06-22 20:55:45

无法映射 Django URL

问题描述

1 个解决方案

解决方案1 1 已采纳 2021-06-22 20:55:45

解决方案1
1 已采纳 2021-06-22 20:55:45