Django管理员内联深层关系 [英] Django admin inline for deep relation

问题描述

内联直接关系很容易，但是如何建立内联以建立更深层的关系呢？

It's quite easy to have inline for direct relation, but how can I create inline for deeper relations?

比方说

class A(models.Model):
     ...

class B(models.Model):
     a = models.ForeignKey(A)

class C(models.Model):
     b = models.ForeignKey(B)

现在我想拥有的是内嵌在 A 的管理员列表中，它将显示所有 C 和相应的 B 名称。我该怎么做？

Now what I'd like to have is inline in admin for A which will show all of it's Cs with corresponding B name. How do I do that?

推荐答案

看看 django-nested-inlines 。例如：

from django.contrib import admin
from nested_inlines.admin import NestedModelAdmin, NestedStackedInline, NestedTabularInline
from models import A, B, C

class MyNestedInline(NestedTabularInline):
    model = C

class MyInline(NestedStackedInline):
    model = B
    inlines = [MyNestedInline,]

class MyAdmin(NestedModelAdmin):
    pass

admin.site.register(A, MyAdmin)

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