因此，我只想在單擊特定文章時輸出其 slug，例如，如果給定文章的 slug 是 django-rules，那么我希望在單擊它時將其輸出為 django-rules。只是這是我的模型
from django.db import models
# Create your models here.
class Article(models.Model):
title = models.CharField(max_length=100)
slug = models.SlugField()
body = models.TextField()
date = models.DateTimeField(auto_now_add = True)
#add in tubnail and author later
def __str__(self):
return self.title
def snippet(self):
return self.body[:50] '...'
這是我的意見.py
from datetime import date
from django.shortcuts import render
from .models import Article
from django.http import HttpRequest
# Create your views here.
def article_list(request):
articles = Article.objects.all().order_by('date')
return render(request,'articles/article_list.html', {'articles': articles} )
def article_detail(request, slug):
return HttpRequest(slug)
網址.py
from posixpath import relpath
from django.contrib import admin
from django.urls import path
from django.views.generic import TemplateView
from . import views
app_name = 'articles'
urlpatterns = [
path('', views.article_list, name = 'list'),
path('<slug:slug>/', views.article_detail, name = 'detail'),
]
請不要建議添加 as_view() 它不起作用。
uj5u.com熱心網友回復：
而不是return HttpRequest(slug)，你需要return HttpResponse(slug)。
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