将网址与网页正确链接的问题

如何解决将网址与网页正确链接的问题

models.py

from django.db import models
from django.core.validators import MaxValueValidator,MinValueValidator
#from django.contrib.postgres.fields import ArrayField

# Create your models here.
class locationData (models.Model):
    locationID = models.AutoField (primary_key = True)
    name = models.CharField (max_length = 64)
    population = models.IntegerField (default = 0,validators = [MinValueValidator(0)])
    apiEndpoint = models.URLField (max_length = 256)
    resourceURL = models.URLField (max_length = 256) 
    
    
class dateData (models.Model):
    entryID = models.AutoField (primary_key = True)
    name = models.CharField (max_length = 64)
    date = models.DateField (auto_Now = False,auto_Now_add = False)
    confirmedCase = models.IntegerField (default = 0,validators = [MinValueValidator(0)])
    deathCase = models.IntegerField (default = 0,validators = [MinValueValidator(0)])

views.py

from django.shortcuts import render
from django.views.generic import TemplateView,ListView,DetailView
from database.models import locationData,dateData

# Create your views here.
class viewLocationData (DetailView):
    template_name = "locationData.html"
    model = locationData
    
    def get_context_data (self,**kwargs):
        location = self.kwargs['location']
        
        context = super().get_context_data (**kwargs)
        context ['location'] = locationData.objects.get (pk = location)
        return context

app/urls.py

from django.urls import path
from database import views

urlpatterns = [
    path ('location_data/<int:location>',views.viewLocationData.as_view(),name = 'location-data')
]

config/urls.py

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

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

locationData.html

<h1>Location Data</h1>

<table>
    <tr>
        <td>Location Name</td>
        <td>{{location.name}}</td>
    </tr>
    <tr>
        <td>Current Estimated Population</td>
        <td>{{location.population}}</td>
    </tr>
    <tr>
        <td>API Endpoint</td>
        <td>{{location.apiEndpoint}}</td>
    </tr>
    <tr>
        <td>URL of Resource</td>
        <td>{{location.resourceURL}}</td>
    </tr>
</table>

我是一名初学者，正在从事一个项目，该项目旨在制作一个基于 Web 的应用程序，该应用程序显示特定位置的 Covid 病例数据。我现在正在尝试制作一个网页，显示有关某个位置的一些简单信息。

此网页的 URL 应该是 http://localhost:8000/database/location_data/1。但是，当我测试该网页时，它无法显示并显示错误消息：

必须使用 URLconf 中的对象 pk 或 slug 调用通用详细信息视图 viewLocationData。

我不确定到底是哪个部分出了问题以及如何修复。

解决方法

您需要传递一个对象标识符 pk 或 slug。

这里的URLconf使用了命名组pk——这个名字是默认的 DetailView 用来查找主键值的名称 过滤查询集。

试试这个

urlpatterns = [
    path ('location_data/<int:pk>',views.viewLocationData.as_view(),name = 'location-data')

    or 

    path ('location_data/<slug:slug>',name = 'location-data')

]

更新：

class viewLocationData (DetailView):
    template_name = "locationData.html"
    model = locationData
    
    def get_context_data (self,**kwargs):
        location = self.kwargs['pk']
        
        context = super().get_context_data (**kwargs)
        context ['location'] = locationData.objects.get (pk = location)
        return context

https://docs.djangoproject.com/en/3.1/ref/class-based-views/generic-display/#detailview