Question

我正在尝试使用多个过滤器查询或不使用

technician = request.POST.get('technician')
## in the previous screen the user may not have informed the parameter, it should not be required

uniorg = request.POST.get('uniorg')
## in the previous screen the user may not have informed the parameter, it should not be required

calls = Call.objects.filter(
    technician=technician,
    uniorg=uniorg,
    ...)

我尝试过：

technician = request.POST.get('technician', '')

和

technician = request.POST.get('technician' or None)

并非所有参数都是必需的。有人可以帮助我吗？

Answer 1

最终密码

list_filters = ['type', 'technician', 'ended', 'uniorg']
filters = {k: v for k, v in request.POST.items() if k in list_filters}

    for l in list_filters:
        if request.POST[l] == '0' or request.POST[l] == '':
            del filters[l]
    calls = Call.objects.filter(start_time__gte=start_date, start_time__lte=end_date, **filters)

Answer 2

您可以尝试这样的事情：

qs = Call.objects.all()

technician = request.POST.get('technician')
if technician is not None:
    qs = qs.filter(technician=technician)

uniorg = request.POST.get('uniorg')
if uniorg is not None:
    qs = qs.filter(uniorg=uniorg)

Answer 3

一种替代方法是使用字典参数扩展：

filters = {k: v for k, v in request.POST.items() if k in ['technician', 'uniorg', ...]}
calls = Call.objects.filter(**filters)

或者，如果您确定POST中只能包含实际的有效过滤器值，则直接传递它：

calls = Call.objects.filter(**request.POST)

Answer 4

您尝试过django-filter吗？这是提供GET参数过滤器的非常好工具（请注意，我不了解您的数据库设计，因此此代码可能无法正常工作）

# filters.py
class CallFilter(django_filters.FilterSet):
    class Meta:
        model = Call
        fields = ['technician', 'uniorg',]

# view.py
items = CallFilter(request.GET, queryset=Call.objects.all())

Django Post或无

4 个答案: