我正在使用 Django 过滤器,并且我希望其中一个字段 ( supervisor
) 成为ChoiceFilter
选择来自模型的对象的位置。最有效的方法是什么?我尝试关注这篇文章,但无论我改变了什么(目前cannot unpack non-iterable int object
),都会不断出错。
# models.py
class people(models.Model):
namelast = models.CharField(max_length=100, verbose_name='Last Name')
namefirst = models.CharField(max_length=100, verbose_name='First Name')
supervisor = models.ForeignKey('self', blank=True, null=True, on_delete=models.SET_NULL, verbose_name='Supervisor')
def __str__(self):
return "%s %s" % (self.namefirst, self.namelast)
# filters.py
class office_filter(django_filters.FilterSet):
supervisor = django_filters.ChoiceFilter(choices=[], lookup_expr='icontains', label='Supervisor')
# other fields
class Meta:
model = people
fields = ['namelast', 'namefirst', 'supervisor']
def __init__(self, *args, **kwargs):
super().__init__(*args, **kwargs)
try:
self.filters['supervisor'].extra['choices'] = [x for x in
people.objects.all().values_list('supervisor', flat=True).distinct()]
except (KeyError, AttributeError):
pass
目标是使该supervisor
字段成为一个很好的菜单,其中包含已添加为people
模型中的主管的所有人员。