2014-09-13 36 views
4

我有查询与此Django的自定义左外连接

news = News.objects.filter(Q(likes__user__isnull=True)|Q(likes__user=user)) 
.extra(select={"is_liked":NewsLikes._meta.db_table+".user_id = %d" % user.id}) 

这给了我下面的查询Django的模型

SELECT (shows_newslikes.user_id = 143) AS `is_liked`, * FROM `shows_news` 
LEFT OUTER JOIN `shows_newslikes` ON (`shows_news`.`id` = `shows_newslikes`.`news_id`) 
WHERE (`shows_newslikes`.`user_id` IS NULL OR `shows_newslikes`.`user_id` = 143) 

我要的是下面的查询的结果

SELECT (shows_newslikes.user_id = 143) AS `is_liked`, * 
FROM `shows_news` LEFT OUTER JOIN `shows_newslikes` ON (`shows_news`.`id` = 
`shows_newslikes`.`news_id` and `shows_newslikes`.`user_id` = 143) WHERE 
(`shows_newslikes`.`user_id` IS NULL ) 

所以我必须做的查询Django模型

+0

,我想这样做没有原始查询 – Muneeb 2014-09-13 08:52:54

+0

究竟什么是你想怎么办?你是否想用一个布尔值来注释每个产生的'News'对象,告诉你该行是否'喜欢'? – 2014-09-13 10:11:19

+0

是的...我想要做的事情,但与左外连接 – Muneeb 2014-09-13 10:50:37

回答

3

如果不使用raw()很难生成此形式的LEFT OUTER JOIN;你也需要distinct()重复的行。我会用EXISTS这是清洁,可能会更快:

news = News.objects.extra(select={'is_liked': 
    'EXISTS (SELECT 1 FROM {tbl_2} ' 
    'WHERE {tbl_2}.news_id = {tbl}.id AND {tbl_2}.user_id = %s)'.format(
     tbl=News._meta.db_table, 
     tbl_2=NewsLikes._meta.dbtable)}, select_params=(user.id,))