Question

#admin.py
class GameListAdmin(admin.ModelAdmin):
    list_display = ['game', 'position']
    ordering = ('position',)

class GameAdmin(admin.ModelAdmin):
    list_display = ['name', 'status']
    actions = [make_published]

#models.py
class Game(models.Model):
    name = models.CharField(max_length=200)
    status = models.CharField(max_length=1, choices=STATUS_CHOICES)

    def __unicode__(self):
        return self.name

class GameList(models.Model):
    game = models.ForeignKey(Game)
    position = models.IntegerField()

    def __unicode__(self):
        return self.game.name

我正在尝试在GameListAdmin的list_display中显示game.status但不是确定如何使用Admin.py中的ForeignKey进行向后查找

有什么想法吗？

Answer 1

你可以这样做：

def get_status(obj):
    return '%s' % (obj.game.status)
get_status.short_description = 'Status'

class GameListAdmin(admin.ModelAdmin):
    list_display = ['game', 'position', get_status]

请参阅docs。

Answer 2

您可以随时定义自己的专栏，例如文档说明：http://docs.djangoproject.com/en/dev/ref/contrib/admin/#django.contrib.admin.ModelAdmin.list_display

例如：

class GameListAdmin(admin.ModelAdmin):
   list_display = ['game_status', 'position']
   ordering = ('position',)

   def game_status(self, obj):
       return obj.game.status
   game_status.short_description = 'Game status'

关于admin.py list_display的问题

2 个答案: