我正在尝试制作它,以便只显示以slug名称为类别的简报,但它不起作用。
目前,我只能通过使用.objects.all((显示所有简报来使用它,但这不适合我想要的用例。
我是否也需要在简短部分中添加段塞字段?
型号.py
class Category(models.Model):
name = models.CharField(max_length=200, unique=True)
slug = models.SlugField(max_length=50, unique=True)
class Meta:
verbose_name_plural = 'categories'
verbose_name = 'category'
def __str__(self):
return self.name
def get_absolute_url(self):
return reverse('browse')
class Brief(models.Model):
author = models.ForeignKey(User, on_delete=models.CASCADE, null=True)
brandname = models.CharField(max_length=28)
description = models.CharField(max_length=200)
date = models.DateTimeField(auto_now=True, blank=True)
category = models.ForeignKey(Category, on_delete=CASCADE)
def get_absolute_url(self):
return reverse('homepage')
URL.py
path('browse/categories/<slug:catslug>/', views.postsinthecategory, name = 'catslug'
views.py
def postsinthecategory(request, catslug):
categories = Category.objects.all()
brief = Brief.objects.all()
if catslug:
category = get_object_or_404(Category, slug = catslug)
brief = Brief.objects.get(category=catslug)
template = 'users/categoryposts.html'
context = {'categories': categories, 'brief': brief, 'category': category}
return render(request, template,context)
您可以使用__slug
来过滤相关模型的slug
字段。使用filter
而不是get
,因为您希望查询集可以包含多个brief
。在您看来,我会将brief
重命名为briefs
,以使其更加清晰。
briefs = Brief.objects.filter(category__slug=catslug)
或者,由于你在前一行提取了类别,你可以这样做:
category = get_object_or_404(Category, slug = catslug)
briefs = Brief.objects.filter(category=category)