【发布时间】:2021-04-03 19:37:55
【问题描述】:
我正在提高我的英语,请耐心等待。
我创建了一个 django-filter 类以在我的仪表板上使用,但 ModelChoiceField 下拉列表显示所有“colocador”对象,而不仅仅是与当前用户相关的对象
Colocador 是与另一个用户个人资料相关的用户个人资料
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 filters.py 中
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']
我将 work_for 硬编码为 3,但显然我不想要硬编码的值,我已经传递了实际用户并尝试了这个但不起作用
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))
下拉列表仅显示与用户相关的对象,但是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,如何创建一个可以正常工作的 ModelChoiceFilter 并只显示与当前用户相关的对象。
【问题讨论】:
标签: django django-filter