Python Django admin-Tablerinline-仅显示其他模型中的一些对象
示例模型:Python Django admin-Tablerinline-仅显示其他模型中的一些对象,python,django,Python,Django,示例模型: class Book(models.Model): TYPES = ( (0, 'Sci-fi') (1, 'Biography') ) title = models.CharField(...) book_type = models.CharField(max_length=1, choices=TYPES) class Connect(models.Model): book1 = models.For
class Book(models.Model):
TYPES = (
(0, 'Sci-fi')
(1, 'Biography')
)
title = models.CharField(...)
book_type = models.CharField(max_length=1, choices=TYPES)
class Connect(models.Model):
book1 = models.ForeignKey(Book, related_name='book_1')
book2 = models.ForeignKey(Book, related_name='book_2')
我想做一个思考,在书中以表格行的形式显示连接,但仅在book1\uuuuu Book\u type=0的地方显示连接
我试着这样做:
class ConnectFormSet(BaseInlineFormSet):
def get_queryset(self):
if not hasattr(self, '_queryset'):
qs = super(ConnectionFormSet, self).get_queryset().filter('book1__book_type':0)
self._queryset = qs
return self._queryset
class InlineConnectn(admin.TabularInline):
fk_name = 'book1'
model = Connect
extra = 0
formset = ConnectFormSet
但它不是我想要的。但是,所有连接在列表行中的连接列表中都可见
Works only save Connect(仅保存与book1的连接。\u book\u type=0)。应执行您想要的操作:
class InlineConnectn(admin.TabularInline):
def formfield_for_foreignkey(self, db_field, request, **kwargs):
if db_field.name == "book1":
kwargs["queryset"] = Foo.objects.filter(book1__book_type=1)
return super(InlineConnectn, self).formfield_for_foreignkey(db_field, request, **kwargs)