Django管理 - 排序list_filter [英] Django admin - sorting list_filter
问题描述
我有'version'在list_filter中,我想要的最新版本是在'全部'列表项后。
在这种情况下,如何排序列表降序? (全部,3.6.99.108,3.6.99.107 ...)
I have 'version' in list_filter, and I want the latest version be after 'All' list item. In this case, how do I sort the list descending? (All, 3.6.99.108,3.6.99.107...)
推荐答案
定义模型中的默认排序 Meta
class:
Define default ordering in model Meta
class:
class MyModel(models.Model):
# some model fields here
class Meta:
ordering = ['-version',]
如果某种方式不符合您的目的,您可以编写自己的自定义列表过滤器:
If somehow it does not fulfill your purpose you can write your own custom list filter:
from django.utils.translation import ugettext_lazy as _
from django.contrib.admin import SimpleListFilter
class VersionFilter(SimpleListFilter):
title = _('version')
parameter_name = 'version'
def lookups(self, request, model_admin):
qs = model_admin.queryset(request)
return [(i, i) for i in qs.values_list('version', flat=True) \
.distinct().order_by('-version')]
def queryset(self, request, queryset):
if self.value():
return queryset.filter(version__exact=self.value())
class MyModel(ModelAdmin):
list_filter = (VersionFilter,)
如果您收到以下错误 AttributeError:object has no属性'queryset'
然后你可能使用 Django 1.5 +
使用 .get_queryset()
相反,例如
If you get following error AttributeError: object has no attribute 'queryset'
then you are probably using Django 1.5+
use .get_queryset()
instead e.g.
qs = model_admin.get_queryset(request)
这篇关于Django管理 - 排序list_filter的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!