django组子类，具有类段塞

Trying to make a linkdirectory page with categories and subcategories, have the following so far, problem is that i can't get the category as a link

尝试用类别和子类别创建一个linkdirectory页面，到目前为止，问题是我不能将类别作为链接。

models.py
class Category(models.Model):
"""Category"""
name = models.CharField(max_length=50)
slug = models.SlugField()

def save(self, *args, **kwargs):

                #self.slug = slugify(self.name)
        self.slug = slugify(self.name)
        super(Category, self).save(*args, **kwargs)
def __unicode__(self):
return self.name


class SubCategory(models.Model):
"""Sub Category"""
category = models.ForeignKey(Category)
name = models.CharField(max_length=50)
slug = models.SlugField()

def save(self, *args, **kwargs):

        self.slug = slugify(self.name)
        super(SubCategory, self).save(*args, **kwargs)

def __unicode__(self):
return self.name

and

和

views.py
@require_GET
def links(request):
"""Linkdirectory Page"""
categories = Category.objects.all()
subcategories = SubCategory.objects.all()
return render(request, 'links.html', {'categories':         categories,             'subcategories': subcategories,})

in order to group the subcategories by category i used the following in

为了将子类别按类别进行分组，我使用了以下方法

{% regroup subcategories|dictsort:"category_id" by category as     categories_list %}

<ul>
{% for categories in categories_list %}
<li>{{ categories.grouper }}
           <li> <a href="/links/{{ categories.slug }}">{{ categories.name     }}</a></li>
<ul>
    {% for item in categories.list %}
            <li><a href="/links/{{ item.slug }}">{{ item.name }}</a></li>
    {% endfor %}
</ul>
</li>
{% endfor %}
</ul>

This all works nicely grouping the subcats together

这些都可以很好地将子猫科动物分组在一起

CAT1 --subcatLINK1 --subcatLINK2

CAT1——subcatLINK1 subcatLINK2

but what i would like to have is

但是我想要的是

CATLINK1 --subcatLINK1 --subcatLINK2

CATLINK1——subcatLINK1 subcatLINK2

Not sure how and if its possible to replace the grouper string with a custom link as like the subcategory link or if i'm going the right way around this

不知道如何以及是否可以用自定义链接替换grouper字符串，就像子类别链接一样，或者我是否按照正确的方式

Edited, adding Bakkals approach below: EDIT AGAIN ;) nevermind, i had to update the naming scheme aswell obviously, below code is the working version. trying to use category.subcategory_set.all as suggested by Bakkal

编辑，添加Bakkals方法:再编辑;)nevermind，我必须更新命名方案，显然，下面的代码是工作版本。尝试使用category.subcategory_set。所有这些都是贝克尔建议的

{% for category in categories %}
<ul>
<li><a href="/links/{{ category.slug }}">{{ category.name }}</a>
{% for subcategory in category.subcategory_set.all %}
<ul>
            <li><a href="/links/{{ subcategory.slug }}">{{     subcategory.name }}</a></li>
</ul>
    {% endfor %}
</li>
</ul>
{% endfor %}
</li>
</ul>
{% endfor %}

1 个解决方案

#1

Since SubCategory contains a ForeignKey to Category, You can use category.subcategory_set to get the subcategories.

因为子类别包含一个类别的外键，你可以使用Category。subcategory_set用来获取子类别。

I find this simpler than manually doing the grouping you did by the category_id

我发现这比手工按category_id进行分组要简单

So you can e.g. pass categories as a queryset of all Category objects as the context to your template, and then within the template

因此，您可以将类别作为所有类别对象的查询集作为上下文传递给模板，然后在模板中传递

{% for category in categories %}
    ...
    {% for subcategory in category.subcategory_set.all %}
    ...
    {% endfor %}
{% endfor %}

#1