如何覆盖list_filter中提供过滤器的查询集?

jul*_*jul 17 django admin

鉴于以下型号

class AnotherModel(models.Model):
    n = models.IntegerField()

class MyModel(models.Model):
    somefield = models.ForeignKey(AnotherModel)
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和管理员

class MyModelAdmin(admin.ModelAdmin):        
    list_filter = ('somefield',)
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如何过滤实例AnotherModel以仅n在管理过滤器中显示具有给定值的实例?

我需要这样的东西:

过滤

通过某个地方

所有

[给定n的AnotherModel实例列表]

Pau*_*ine 23

请参见ModelAdmin.querysetModelAdmin.formfield_for_foreignkey.来自文档:

ModelAdmin上的queryset方法返回可由管理站点编辑的所有模型实例的QuerySet.覆盖此方法的一个用例是显示登录用户拥有的对象:

class MyModelAdmin(admin.ModelAdmin):
    def queryset(self, request):
        qs = super(MyModelAdmin, self).queryset(request)
        if request.user.is_superuser:
            return qs
        return qs.filter(author=request.user)
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ModelAdmin上的formfield_for_foreignkey方法允许您覆盖外键字段的默认表单字段.例如,要根据用户返回此外键字段的对象子集:

class MyModelAdmin(admin.ModelAdmin):
    def formfield_for_foreignkey(self, db_field, request, **kwargs):
        if db_field.name == "car":
            kwargs["queryset"] = Car.objects.filter(owner=request.user)
        return super(MyModelAdmin, self).formfield_for_foreignkey(db_field, request, **kwargs)
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这使用HttpRequest实例过滤Car外键字段,以仅显示User实例拥有的汽车.

[更新]

对不起,我没看过"过滤器"部分.在Django> = 1.4,你可以传递的一个子类 django.contrib.admin.SimpleListFilterlist_filter参数列表,你可以以覆盖查找和查询集方法使用.

from datetime import date

from django.contrib import admin
from django.utils.translation import ugettext_lazy as _

class DecadeBornListFilter(admin.SimpleListFilter):
    # Human-readable title which will be displayed in the
    # right admin sidebar just above the filter options.
    title = _('decade born')

    # Parameter for the filter that will be used in the URL query.
    parameter_name = 'decade'

    def lookups(self, request, model_admin):
        """
        Returns a list of tuples. The first element in each
        tuple is the coded value for the option that will
        appear in the URL query. The second element is the
        human-readable name for the option that will appear
        in the right sidebar.
        """
        return (
            ('80s', _('in the eighties')),
            ('90s', _('in the nineties')),
        )

    def queryset(self, request, queryset):
        """
        Returns the filtered queryset based on the value
        provided in the query string and retrievable via
        `self.value()`.
        """
        # Compare the requested value (either '80s' or '90s')
        # to decide how to filter the queryset.
        if self.value() == '80s':
            return queryset.filter(birthday__gte=date(1980, 1, 1),
                                birthday__lte=date(1989, 12, 31))
        if self.value() == '90s':
            return queryset.filter(birthday__gte=date(1990, 1, 1),
                                birthday__lte=date(1999, 12, 31))

class PersonAdmin(admin.ModelAdmin):
    list_filter = (DecadeBornListFilter,)
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  • 使用 Django 1.10,您需要覆盖 get_queryset() 而不是 queryset()。 (2认同)

sed*_*nym 7

编辑 - 已指出此方法存在问题,请参阅下文

你可以这样做:

假设你有一个名为Animal的模型,它有一个名为Species的模型的ForeignKey字段.在特定的管理列表中,您希望仅允许在动物过滤器选项中显示某些物种.

首先,在Animal的ModelAdmin中指定一个名为SpeciesFilter的自定义ListFilter:

class AnimalAdmin(ModelAdmin):
    list_filter = (('species', SpeciesFilter),)
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然后定义SpeciesFilter:

from django.contrib.admin.filters import RelatedFieldListFilter

class SpeciesFilter(RelatedFieldListFilter):
    def __init__(self, field, request, *args, **kwargs):
        """Get the species you want to limit it to.
        This could be determined by the request,
        But in this example we'll just specify an
        arbitrary species"""
        species = Species.objects.get(name='Tarantula')

        #Limit the choices on the field
        field.rel.limit_choices_to = {'species': species}

        #Let the RelatedFieldListFilter do its magic 
        super(SpeciesFilter, self).__init__(field, request, *args, **kwargs)
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应该这样做.

  • 这可能会导致意外结果!(我用Django 1.5尝试过.)过滤器查找以某种方式缓存.因此``RelatedFieldListFilter`不适用于动态查找(例如,如果它们在用户之间变化).所有用户都将获得相同的过滤器.[文件](https://docs.djangoproject.com/en/dev/ref/contrib/admin/#django.contrib.admin.ModelAdmin.list_filter)"MiniListFilter`没有这样的问题.有关如何操作,请参阅[此答案](http://stackoverflow.com/a/22708389/2704544). (2认同)