本人有起诉人名单,而且想将案件定性为不敏感案件。
members = Person.objects.filter(person=person_id)
members_list = list(members)
members_list.sort( key = operator.attrgetter( sort_by ), reverse = False )
here sort_by have attribute name by whom i want to sort. Now how can i sort with case insensitive ??
请提供帮助。
如果sort_by是一个变数,含有你希望用于分类的属性名称的缩略语,而且你希望使用operator.attrgetter(<><>>>/code>,你可以使用这一变量,将属性的价值转换到所有较低类别,以便进行比较:
members_list.sort( key = lambda mbr: operator.attrgetter( sort_by )( mbr ).lower(),
reverse = False )
虽然如下一些内容更容易读写并更有效率:
get_key = operator.attrgetter( sort_by )
members_list.sort( key = lambda mbr: get_key( mbr ).lower(), reverse = False )
members_list.sort( key = lambda member: getattr(member,sort_by).lower(), reverse = False )
看来,这个问题涉及jan。 你们可能会想回到一个分类的问候。 这意味着你可以离开数据库,而不是自发。 见网站:django。
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