Django管理员内联深层关系

var*_*tec 6 django django-admin

内联直接关系很容易,但是如何为更深层次的关系创建内联?

让我们说情况就像

class A(models.Model):
     ...

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

class C(models.Model):
     b = models.ForeignKey(B)
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现在我想要的是内联管理员A,它将显示所有C相应B名称的内容.我怎么做?

sta*_*alk 4

看一下django-nested-inlines。从例子:

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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