Django - 用户管理员 - 将组添加到list_display

Question

众所周知,ManytoMany关系不能列在list_display中.是否有任何工作可以使它像group1,group2等？

Answer 1

奖励:将组显示为用户过滤器:

class SBUserAdmin(UserAdmin):
    list_filter = UserAdmin.list_filter + ('groups__name',)
    list_display = ('username','custom_group', )

    def custom_group(self, obj):
        """
        get group, separate by comma, and display empty string if user has no group
        """
        return ','.join([g.name for g in obj.groups.all()]) if obj.groups.count() else ''

admin.site.unregister(User)
admin.site.register(User, SBUserAdmin)

Answer 2

我不理解您的示例（group1，group2），但是您可以在更改列表视图中将任何函数用作列，这意味着您很可能可以执行想要显示的操作！

例：

class MyModelAdmin(admin.ModelAdmin):
    list_display = ('foo', 'bar')

    def foo(self):
        return "This column is Foo"

    def bar(self, obj):
        try:
            return obj.m2m.latest('id')
        except obj.DoesNotExist:
            return "n/a"


    # there's a few more things you can do to customize this output
    def bar(self, obj):
        return '<span style="color:red;">By the way, I am red.</span>'

    bar.short_description = "My New Column Label"
    bar.allow_tags = True