在 Django 管理中为 list_filter 创建自定义过滤器

我想为 django 管理员制作自定义过滤器，而不是通常的“ is _ staff”和“ is _ superuser”。我在姜戈的文档中读过这个 List _ filter。 自定义过滤器是这样工作的:

from datetime import date


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


class DecadeBornListFilter(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(ModelAdmin):
list_filter = (DecadeBornListFilter,)

但是我已经为 list _ display 定制了如下函数:

def Student_Country(self, obj):
return '%s' % obj.country
Student_Country.short_description = 'Student-Country'

Is it possible i could use the custom functions for list_display in list_filter instead of writing a new custom function for list_filter? Any suggestions or improvements are welcome.. Need some guidance on this... Thanks...

79143 次浏览
小开

您的 list_display方法返回一个字符串，但如果我理解正确，您想要做的是添加一个过滤器，允许选择国家的学生，对吗？

对于这个简单的关系过滤器，实际上对于“ Student-Country”列表显示列，您不需要创建自定义过滤器类，也不需要创建自定义列表显示方法; 这就足够了:

class MyAdmin(admin.ModelAdmin):
list_display = ('country', )
list_filter = ('country', )

Django 执行 list_filter的方式，如解释的 在文件里所示，首先是自动匹配您提供给预构建过滤器类的字段; 这些过滤器包括 CharField 和 ForeignKey。

类似地，list_display通过检索相关对象并返回这些对象的 Unicode值(与上面提供的方法相同) ，使用传递的字段自动填充变更列表列。

小开

您确实可以通过扩展 SimpleListFilter 向管理筛选器添加自定义筛选器。例如，如果你想为“ Africa”添加一个大陆过滤器到上面使用的国家管理过滤器中，你可以执行以下操作:

在 admin.py 中:

from django.contrib.admin import SimpleListFilter


class CountryFilter(SimpleListFilter):
title = 'country' # or use _('country') for translated title
parameter_name = 'country'


def lookups(self, request, model_admin):
countries = set([c.country for c in model_admin.model.objects.all()])
return [(c.id, c.name) for c in countries] + [
('AFRICA', 'AFRICA - ALL')]


def queryset(self, request, queryset):
if self.value() == 'AFRICA':
return queryset.filter(country__continent='Africa')
if self.value():
return queryset.filter(country__id__exact=self.value())


class CityAdmin(ModelAdmin):
list_filter = (CountryFilter,)
小开

除了 Rick Westera的答案，这里是这种情况下的 姜戈医生

ModelAdmin.list_filter
设置 list_filter以激活管理员更改列表页面右侧边栏中的过滤器
list_filter应该是一个元素列表或元组


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