Question

我正在使用Django建立一个网站。我想结合多个过滤结果（Querysets）。

我模特的关系

'工作人员'1：m'会员'1：m'PaymentHistory'1：1'RefundHistory'

我的view.py：

wanted_refund = set()
for m in staff.members.all():
    payment = m.PaymentHistory.filter(division="Membership")
    for p in payment:
        try:
            refund = RefundHistory.objects.filter(payment=p).filter(refund_date__range=[this_month_start, date])

            wanted_refund.add(refund)
        except RefundHistory.DoesNotExist:
            pass
 context = {
             'wanted_refund' : wanted_refund,}
    return render(request, 'refund.html', context)

但是，使用过滤器无效。它只适用于我使用'get'。

print(refund)向我展示了这样的结果：

 < QuerySet [] >

 < QuerySet [] >

 < QuerySet [] >

 < QuerySet [< RefundHistory: RefundHistory object >] >

我想只使用具有该对象的Querysets，我想要的是模板中的下面一个：

{% for refund in wanted_history %}
 {{ refund.refund_date }}
 {{ refund.refund_amount}}
 {% endfor %}

如何将多个过滤器结果传递给for循环？

Answer 1

您可以使用管道运算符加入查询集：

wanted_refund = RefundHistory.objects.none()
for m in staff.members.all():
    payment = m.PaymentHistory.filter(division="Membership")
    for p in payment:
        try:
            wanted_refund |= RefundHistory.objects.filter(payment=p).filter(refund_date__range=[this_month_start, date])
        except RefundHistory.DoesNotExist:
            pass
wanted_refund = wanted_refund.distinct()
context = {'wanted_refund': wanted_refund}
return render(request, 'refund.html', context)

此外，如果它适合您，您可以使用wanted_refund.update(list(refund))代替wanted_refund.add(refund)

Answer 2

使用RefundHistory的单个查询，可以获得每位会员每次付款的退款记录：

payments = []
for m in staff.members.all():
    payments.extend(m.PaymentHistory.filter(division="Membership").values_list('pk', flat=True))

wanted_refund = RefundHistory.objects.filter(payment__pk__in=payments, refund_date__range=[this_month_start, date])

context = {'wanted_refund' : wanted_refund,}
return render(request, 'refund.html', context)

将过滤器结果合并为for循环

2 个答案: