关于用网页正确制作 link 一个 URL 的问题

Question

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：

Generic detail view viewLocationData must be called with either an object pk or a slug in the URLconf.

我不确定到底是哪一部分出了问题以及如何解决它。

Answer 1

您需要传递对象标识符 pk 或 slug。

The URLconf here uses the named group pk - this name is the default name that DetailView uses to find the value of the primary key used to filter the queryset.

试试这个

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

    or 

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

]

已更新：

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

关于用网页正确制作 link 一个 URL 的问题

Problem about making correctly link a URL with a webpage

python

django

django-views

django-urls