Question

我的SQL性能很差，其中包含一个不相关的子查询。我用一个共同相关的子查询替换它，但性能变得更糟。有没有办法通过完全删除子查询并用连接替换它来重写此SQL？

以下是查询的简化版本：

    select distinct tab1.app_id, 
            tab1.name, 
            tab1.stat_cd, 
            tab1.qr 
    from apps tab1 , issues tab2 
    where 
            tab1.app_id = tab2.app_id and 
            tab1.qr = 'm' and 
            tab2.iqr = 'm' and 
            tab1.app_id not in 
            ( 
              select  distinct tab3.app_id 
              from issues tab3 
              where tab3.iqr = 'm' and 
              ( 
                 tab3.i_cd = 'f' or 
                 tab3.i_cd = 'r' or 
                 tab3.i_cd = 'c' 
              ) 
            )

任何提示或帮助表示赞赏。谢谢。

Answer 1

不确定这会有多大帮助，但我想听听它是如何做到的......

SELECT DISTINCT tab1.app_id,
       tab1.name, 
       tab1.stat_cd, 
       tab1.qr
FROM apps tab1
JOIN issues tab2 ON tab1.app_id = tab2.app_id
JOIN ( SELECT DISTINCT app_id AS app_id
       FROM issues
       WHERE iqr = 'm'
         AND ( i_cd = 'f' OR
               i_cd = 'r' OR
               i_cd = 'c'
             ) 
     ) tab3 ON tab1.app_id = tab3.app_id
WHERE tab1.qr = 'm'
  AND tab2.iqr = 'm';

Answer 2

试试这个： -

 select distinct tab1.app_id, 
            tab1.name, 
            tab1.stat_cd, 
            tab1.qr 
     from apps tab1
     inner join
     issues tab2 
     on 
     tab1.app_id = tab2.app_id
     left join
     issues tab3 
     on
     tab1.app_id = tab3.app_id
  where     tab1.qr = 'm' and 
            tab2.iqr = 'm' and
            concat(tab3.iqr,tab3.i_cd) not in ('mf','mr','mc)

希望这会有所帮助： - ）

Answer 3

试试这个

select distinct tab1.app_id, 
        tab1.name, 
        tab1.stat_cd, 
        tab1.qr 
from apps tab1 
inner join issues tab2 on ( tab1.app_id, tab1.qr)= ( tab2.app_id, tab2.iqr) and tab1.qr='m'
where not exists 
( 
          select  * from issues tab3 
          where ( tab1.app_id, tab1.qr)= ( tab3.app_id, tab3.iqr)
          and  tab3.i_cd in ('f', 'r', 'c')
)

Answer 4

其他解决方案

select distinct tab1.app_id, 
        tab1.name, 
        tab1.stat_cd, 
        tab1.qr 
from apps tab1 
inner join issues tab2 on ( tab1.app_id, tab1.qr)= ( tab2.app_id, tab2.iqr) and tab1.qr='m'
left outer join issues tab3 on ( tab1.app_id, tab1.qr)= ( tab3.app_id, tab3.iqr) and tab3.i_cd in ('f', 'r', 'c') 
where tab3.app_id is null

删除子查询所需的SQL帮助

4 个答案: