在Django中,我有一个视图,在该视图中,我想按非主键字段进行分组,然后为每个组计数满足谓词的行数。我试过的代码是计算在Django中满足谓词的行
funding_requests=FRNs.objects.filter(spin=spin).extra(
select={"approved": "committed_amount > 0",
"denied": "committed_amount=0"}
).values("funding_year").annotate(
requested=Sum('orig_commitment_request'), approved=Sum('committed_amount'),
disbursed=Sum('total_authorized_disbursement'), num_requests=Count("funding_year"),
num_approved=Sum('approved'), num_denied=Sum('denied')
)
但我得到错误消息“Can not resolve keyword'denied'into field。”我也试过
funding_requests=FRNs.objects.filter(spin=spin).values("funding_year").annotate(
requested=Sum('orig_commitment_request'), approved=Sum('committed_amount'),
disbursed=Sum('total_authorized_disbursement'), num_requests=Count("funding_year")
).extra(
select={"num_approved": "sum(committed_amount > 0)",
"num_denied": "sum(committed_amount=0)"}
)
,我没有得到一个错误,但num_approved
和num_denied
显示在页面上为空白。有谁知道如何获得满足给定谓词的每个分组的行数。
我有这种说法,这样的数额将每年的例子是https计算,如: //docs.djangoproject.com/en/1.3/topics/db/aggregation/#values – murgatroid99