您可以执行以下操作：

py（示例）：

 模型B（models.Model）：
 name = models.CharField（max_length = 20 ）

模型A（models.Model）：
 field1 = models.CharField（max_length = 20）
 Bkey = models.ForeignKey（B）
 

admin.py

  from django.core import urlresolvers 

 class AAdmin（admin.ModelAdmin）：
 list_display = [field1，link_to_B] 
 def link_to_B（self，obj）：
 link = urlresolvers.reverse（admin：yourapp_b_change，args = [obj.B.id]）#model name必须为小写
 return u'< a href =％s>％ s< / a>'％（link，obj.B.name）
 link_to_B.allow_tags = True 
 

将yourapp替换为您的应用程序的名称。

I have a model A with a ForeignKey to a model B. In Django admin, how can I add a link in the admin page of model A next to the ForeignKey field which open the admin page of the model B ?

You can do the following:

models.py (example):

model B(models.Model):
    name = models.CharField(max_length=20)

model A(models.Model):
    field1 = models.CharField(max_length=20)
    Bkey = models.ForeignKey(B)

admin.py

from django.core import urlresolvers

class AAdmin(admin.ModelAdmin):
    list_display = ["field1","link_to_B"]
    def link_to_B(self, obj):
        link=urlresolvers.reverse("admin:yourapp_b_change", args=[obj.B.id]) #model name has to be lowercase
        return u'<a href="%s">%s</a>' % (link,obj.B.name)
    link_to_B.allow_tags=True

Replace yourapp with the name of your app.