django - Django Admin 中的通用外键

Question

是否可以通过 GenericForeignKey 过滤Django 管理中的对象标题？

我想按程序名称过滤，要么NonSupportedProgram.title或 SupportedProgram.title ( list_filter = (SOME FIELD HERE) )，但不知道怎么做？

模型.py

class FullCitation(models.Model):
    # the software to which this citation belongs
    # either a supported software program or a non-supported software program
    content_type = models.ForeignKey(ContentType)
    object_id = models.PositiveIntegerField()
    content_object = generic.GenericForeignKey('content_type', 'object_id')

class NonSupportedProgram(models.Model):
    title = models.CharField(max_length=256, blank = True)
    full_citation = generic.GenericRelation('FullCitation')

class SupportedProgram(models.Model):
    title = models.CharField(max_length=256, blank = True)
    full_citation = generic.GenericRelation('FullCitation')

Answer 1

也许您可以使用 SimpleListFilter 并定义自己的查询集，以便获得自己的结果。

更多信息在这里:
https://docs.djangoproject.com/en/1.8/ref/contrib/admin/#django.contrib.admin.ModelAdmin.list_filter

class YourCustomListFilter(admin.SimpleListFilter):
   title = 'By Program Name'
   parameter_name = 'program'

   def lookups(self, request, model_admin):
       return(
           ('supported','Supported Programs'),
           ('nonsupported', 'Non-Supported Programs')
       )

   def queryset(self, request, queryset):
        if self.value() == 'supported':
            return queryset.filter(supported__isnull=False)
        if self.value() == 'nonsupported':
            return queryset.filter(nonsupported__isnull=False)

admin.py
list_filter = (YourCustomListFilter,)

您需要在 GenericRelation 声明中添加 related_query_name='supported' 和 related_query_name='nonsupported' 以允许从相关对象进行查询。

有关在此处查询 GenericRelations 的更多信息:
https://docs.djangoproject.com/en/1.8/ref/contrib/contenttypes/#reverse-generic-relations

这应该是对正确方向的一点插入。