Question

我有一个返回数值的模型函数。我希望能够按该值进行过滤和排序。当我尝试将其添加到list_filter时，Django抱怨该模型没有这样的字段。我可以告诉Django将它视为一个领域吗？如果是这样，怎么样？

Answer 1

你是否曾尝试将它用作类属性，但我没有尝试过它？

class SomeModel(models.Model):
    @property
    def list_filter(self):
        """ Do your stuffs here """

Answer 2

您可以编写自定义过滤器并将其传递到list_filter：

from django.contrib.admin import SimpleListFilter

class CustomFilter(SimpleListFilter):
    # Human-readable title which will be displayed in the
    # right admin sidebar just above the filter options.
    title = _('active status')

    # Parameter for the filter that will be used in the URL query.
    parameter_name = 'status'

    def lookups(self, request, model_admin):
        if request.user.is_superuser:
            return (
                ('active', _('Active')),
                ('not_active', _('Not Active')),
            )

    def queryset(self, request, queryset):
        # do something here with queryset

class MyAdmin(admin.ModelAdmin):
    list_filter = (CustomFilter, 'other_model_field')

有关ModelAdmin.list_filter

的详情，请参阅here

我可以在管理员中使用Django模型函数作为字段吗？

2 个答案: