Django 管理员 : Setting list_display conditionally

Question

是否有任何管理模型方法，如 get_list_display() 或某种方式，我可以有一些条件来设置不同的 list_display 值？

class FooAdmin (model.ModelAdmin):
    # ...
    def get_list_display ():
        if some_cond:
            return ('field', 'tuple',)
        return ('other', 'field', 'tuple',)

Answer 1

ModelAdmin 类有一个名为 get_list_display 的方法，该方法将 request 作为参数，默认返回该类的 list_display 属性。

所以你可以这样做:

class ShowEFilter(SimpleListFilter):
    """ A dummy filter which just adds a filter option to show the E column,
        but doesn't modify the queryset.
    """
    title = _("Show E column")
    parameter_name = "show_e"

    def lookups(self, request, model_admin):
        return [
            ("yes", "Yes"),
        ]

    def queryset(self, request, queryset):
        return queryset

class SomeModelAdmin(admin.ModelAdmin):

    list_display = (
      "a",
      "b",
      "c",
      "d",
      "e"
    )
    list_filter = (
        ShowEFilter,
    )


    def get_list_display(self, request):
        """ Removes the E column unless "Yes" has been selected in the 
            dummy filter.
        """
        list_display = list(self.list_display)
        if request.GET.get("show_e", "no") != "yes":
            list_display.remove("e")

        return list_display