Question

我需要什么

我需要在mysql中对整数进行排序。

取消订阅tinyint（6）

1,1,4,5,1,1，......等

sql查询

 SELECT e0_.metadata AS metadata0, u1_.unsubscribe AS unsubscribe1, u1_.id AS id2, u1_.linkedin_id AS linkedin_id3, u1_.facebook_id AS facebook_id4, u1_.linkedin_profile AS linkedin_profile5, u1_.profile AS profile6, 
    CASE WHEN u1_.linkedin_id <> ''
    OR u1_.facebook_id <> ''
    THEN 1 
    ELSE 0 
    END AS sclr7, 
    CASE WHEN e0_.edition =23932
    THEN 'current'
    ELSE 'past'
    END AS sclr8
    FROM event_visitor e0_, user u1_
    WHERE e0_.user = u1_.id
    AND e0_.event =25162
    AND e0_.published =1
    AND e0_.showme =1
    GROUP BY e0_.user
    ORDER BY sclr7 DESC , e0_.id DESC , LENGTH( u1_.unsubscribe ) 
    LIMIT 30

我试过的解决方案。
1. 长度
2. cast
3. 按col * 1，col

但没有解决方案在tiny int数据类型中工作。

问题

还有像-5这样的负数。

o / p应该是

all 1 should be at top .

最后经过谷歌搜索

按照CAST命令（est_dily_pple as SIGNED INTEGER）ASC
按顺序排序（u1_.unsubscribe AS SIGNED INTEGER）ASC，sclr7 DESC，e0_.id DESC
- 但问题o / p -5,1,1，...等等。

解决方案我出现了：

ORDER BY SIGN（CAST（u1_.unsubscribe AS SIGNED INTEGER））ASC，ABS（CAST（u1_.unsubscribe AS SIGNED INTEGER）），sclr7 DESC，e0_.id DESC。

Refrence

  http://stackoverflow.com/questions/11282571/specific-order-by-in-mysql-statement

我需要实现abs函数学说

dql代码

     $from = 'Entities\EventVisitor visitor,Entities\User user';
   $qb = $this->em->createQueryBuilder();

  $qb->select("visitor.metadata,ABS( u1_.unsubscribe )as unsubscribe)
 ->add('from',$from)

 ->where('visitor.user=user.id')
 ->groupBy('visitor.user')
->orderBy('unsubscribe','asc')

error 500 internal error.

如何在mysql中对数据类型tinyint进行排序

0 个答案: