自定义标签，例如使用 django-filter

Question

我有观察者模型，它通过 OneToOneRelation 扩展用户模型。我对具有观察者外键的模型清单进行了过滤。这是 models.py 代码：

class Observer(models.Model):
    user = models.OneToOneField(User, on_delete=models.CASCADE)
    rating = models.IntegerField(default=0)
    bird_ratings = models.ManyToManyField('BirdName', through='BirdRatings')

class Checklist(gismodels.Model):
    date = models.DateField()
    time_start = models.TimeField(null=True)
    time_end = models.TimeField(null=True)
    site_name = models.CharField(max_length=200)
    municipality = models.CharField(max_length=200)
    observer = models.ForeignKey(Observer, null=True, default='')
    other_observers = models.CharField(max_length=150, default='')
    note = models.CharField(max_length=2000)
    location_note = models.CharField(max_length=2000, default='')
    position = gismodels.PointField(null=True, srid=4326)
    created = models.DateTimeField(db_index=True, null=True)
    rating = models.IntegerField(default=0)

然后我在filters.py中制作了ChecklistFilter类：

class ChecklistFilter(django_filters.FilterSet):
    date = DateFromToRangeFilter(label='Datum pozorování', widget=RangeWidget(attrs={'display': 'inline'}))
    #groups = django_filters.ModelMultipleChoiceFilter(queryset=User.objects.filter(), widget=forms.CheckboxSelectMultiple)
    created = DateFilter(label='Datum vložení')
    observer = ModelChoiceFilter(queryset=Observer.objects.filter(), label='Pozorovatel')
    municipality = CharFilter(label='Obec')
    oblast = CharFilter(label='Oblast')
    rating = NumberFilter(label='Hodnocení')

    class Meta:
        model = Checklist
        fields = ['date', 'created', 'observer', 'municipality', 'site_name', 'rating']

这是我的模板代码的一部分，我在其中使用观察者字段：

<div class="col-sm-2">
    {{ filter.form.observer.label_tag }}
    {% render_field filter.form.observer class="form-control" %}
</div>

但问题是我不需要获取观察者对象，但我需要来自用户模型的last_name，它链接到观察者模型。在表单中，我只是扩展 ModelChoiceField 并覆盖 label_from_instance() 方法。 但这在这里不起作用。 Django-filter 使用此类，但覆盖在这里不起作用。我试过这个：

from .models import Checklist, Observer
from django.contrib.auth.models import User
import django_filters
from django_filters import DateFromToRangeFilter, DateFilter, CharFilter, NumberFilter, ModelChoiceFilter
from django_filters.widgets import RangeWidget
from django import forms


class ChecklistModelChoiceField(forms.ModelChoiceField):
    def label_from_instance(self, obj):
        return obj.user.last_name

class ChecklistFilter(django_filters.FilterSet):
    date = DateFromToRangeFilter(label='Datum pozorování', widget=RangeWidget(attrs={'display': 'inline'}))
    #groups = django_filters.ModelMultipleChoiceFilter(queryset=User.objects.filter(), widget=forms.CheckboxSelectMultiple)
    created = DateFilter(label='Datum vložení')
    observer = ChecklistModelChoiceField(queryset=Observer.objects.filter(), label='Pozorovatel')
    municipality = CharFilter(label='Obec')
    oblast = CharFilter(label='Oblast')
    rating = NumberFilter(label='Hodnocení')

    class Meta:
        model = Checklist
        fields = ['date', 'created', 'observer', 'municipality', 'site_name', 'rating']

网站上的选择字段有观察者对象：

编辑
是的，我都尝试过。然后我尝试了：

class ChecklistModelChoiceField(django_filters.ModelChoiceFilter):
    def label_from_instance(self, obj):
        return obj.user.last_name

没有错误，但是没有使用

label_from_instance()

。仍然是 select 中的 Observer 对象。

我犯了一个错误。 ModelChoiceFilter 不扩展 ChoiceField。过滤器中的代码如下所示。这只是财产：

class ModelChoiceFilter(QuerySetRequestMixin, Filter):
    field_class = forms.ModelChoiceField

所以我的第一个想法行不通。还有其他机会如何制作我需要的东西吗？

Answer 1

我知道这个问题已经存在一年多了，但我遇到了同样的问题并使用 ModelChoiceFilter 找到了解决方案。所以如果其他人遇到这个：

from .models import Checklist, Observer
from django.contrib.auth.models import User
import django_filters
from django_filters import DateFromToRangeFilter, DateFilter, CharFilter, NumberFilter, ModelChoiceFilter
from django_filters.widgets import RangeWidget
from django import forms

class ChecklistFilter(django_filters.FilterSet):
    date = DateFromToRangeFilter(label='Datum pozorování', widget=RangeWidget(attrs={'display': 'inline'}))
    #groups = django_filters.ModelMultipleChoiceFilter(queryset=User.objects.filter(), widget=forms.CheckboxSelectMultiple)
    created = DateFilter(label='Datum vložení')
    observer = ModelChoiceFilter(queryset=Observer.objects.filter(), label='Pozorovatel')
    municipality = CharFilter(label='Obec')
    oblast = CharFilter(label='Oblast')
    rating = NumberFilter(label='Hodnocení')

    class Meta:
        model = Checklist
        fields = ['date', 'created', 'observer', 'municipality', 'site_name', 'rating']

    def __init__(self, *args, **kwargs):
        super(ChecklistFilter, self).__init__(*args, **kwargs)
        # You need to override the label_from_instance method in the filter's form field
        self.filters['thon_group'].field.label_from_instance = lambda obj: obj.user.last_name

Answer 2

我找到了解决方案。我使用 ChoiceFilter 而不是 ModelChoiceFilter，然后对于选择调用一个方法，在该方法中我使用观察者 id 和用户模型中的名称创建元组的元组。

def get_last_names():
    last_names = ()
    observers = Observer.objects.all()
    for obs in observers:
        last_names += (obs.id, obs.user.last_name),
    return last_names

observer = ChoiceFilter(choices=get_last_names, label='Observer')

Answer 3

没有一个答案对我有用，所以这是一个：

from django import forms

class FullNameUserChoiceField(forms.ModelChoiceField):
    def label_from_instance(self, obj):
        return f"{obj.last_name}, {obj.first_name}"

class FullNameUserChoiceFilter(django_filters.ModelChoiceFilter):
    field_class = FullNameUserChoiceField

class MyModelFilter(django_filters.FilterSet):
    instructors = User.objects.filter(groups__name='Department_A').order_by(Lower('last_name'))
    instructor = FullNameUserChoiceFilter(queryset=instructors)

    class Meta:
        model = MyModel
        fields = {
             ...