django过滤器上的ModelChoiceFilter

django过滤器上的ModelChoiceFilter,django,django-filter,Django,Django Filter,我正在提高我的英语水平,请耐心点 我创建了一个django过滤器类用于我的仪表板,但是ModelChoiceField下拉列表显示了所有“colocador”对象,而不仅仅是与当前用户相关的对象 Colocador是与上的另一个用户配置文件相关的用户配置文件 class Colocador(models.Model): user = models.OneToOneField( MyUser, on_delete=models.CASCADE, primary_key=Tr

我正在提高我的英语水平,请耐心点

我创建了一个django过滤器类用于我的仪表板,但是ModelChoiceField下拉列表显示了所有“colocador”对象,而不仅仅是与当前用户相关的对象

Colocador是与上的另一个用户配置文件相关的用户配置文件

class Colocador(models.Model):
    user = models.OneToOneField(
        MyUser, on_delete=models.CASCADE, primary_key=True)
    work_for = models.ForeignKey(Dono, models.CASCADE, blank=True, null=True)
我有一个这样的模型

class Venda(models.Model):
    name = models.CharField(max_length=50, verbose_name='Nome')
    colocador = models.ForeignKey(
        Colocador, on_delete=models.CASCADE, verbose_name="Colocador")
    
在我的过滤器里

class VendaFilter(django_filters.FilterSet):
    foo
    colocador = django_filters.ModelChoiceFilter(
        queryset=Colocador.objects.filter(work_for=3))

    class Meta:
        model = Venda
        fields = ['search', 'status', 'colocador']
我硬编码了work_为3,但显然我不想要硬编码的值,我已经传递了实际的用户并尝试了这个方法,但没有成功

class VendaFilter(django_filters.FilterSet):
    search = django_filters.CharFilter(
        method='search_filter', label='Procurar')
    # colocador = django_filters.ModelChoiceFilter(
    #     queryset=Colocador.objects.filter(work_for=3))

    class Meta:
        model = Venda
        fields = ['search', 'status', 'colocador']

    def search_filter(self, queryset, name, value):
        return Venda.objects.filter(
            Q(name__icontains=value) | Q(phone_number__icontains=value) | Q(
                cpf__icontains=value) | Q(rg__icontains=value), colocador__in=self.colocadores)

    def __init__(self, user, *args, **kwargs):
        super(VendaFilter, self).__init__(*args, **kwargs)
        self.colocadores = Colocador.objects.filter(work_for=user)
        self.filters['colocador'] = django_filters.ModelChoiceFilter(
            queryset=Colocador.objects.filter(work_for=user))
下拉列表只显示与用户相关的对象,但当我提交过滤器时会引发此错误

raise FieldError("Cannot resolve keyword '%s' into field. "
django.core.exceptions.FieldError: Cannot resolve keyword 'None' into field. Choices are: acertado, colocador, colocador_id, cpf, expiration_date, id, mercadoria, mercadoria_id, name, phone_number, rg, sale_date, status

那么,如何创建一个正常工作的ModelChoiceFilter,并向我显示与当前用户相关的对象。

如果有人有同样的问题,这里是解决方案

def __init__(self, user, *args, **kwargs):
    super(VendaFilter, self).__init__(*args, **kwargs)
    self.colocadores = Colocador.objects.filter(work_for=user)
    self.filters['colocador'].queryset = Colocador.objects.filter(
        work_for=user)
简单而美丽