更新时间:2024-07-03 GMT+08:00
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案例:改写SQL消除in-clause

现象描述

in-clause/any-clause是常见的SQL语句约束条件,有时in或any后面的clause都是常量,类似于:

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select count(1) from calc_empfyc_c1_result_tmp_t1 where ls_pid_cusr1 in (20120405, 20130405);

或者

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select count(1) from calc_empfyc_c1_result_tmp_t1 where ls_pid_cusr1 in any(20120405, 20130405);

但是也有一些如下的特殊用法:

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SELECT ls_pid_cusr1,COALESCE(max(round((current_date-bthdate)/365)),0)FROM calc_empfyc_c1_result_tmp_t1 t1,p10_md_tmp_t2 t2WHERE t1.ls_pid_cusr1 = any(values(id),(id15))GROUP BY ls_pid_cusr1;

其中,id、id15为p10_md_tmp_t2中的两列,“t1.ls_pid_cusr1 = any(values(id),(id15))”等价于“t1.ls_pid_cusr1 = id or t1.ls_pid_cusr1 = id15”

因此join-condition实质上是一个不等式,这种非等值的join操作必须使用nestloop连接,对应执行计划如下:

优化说明

测试发现由于两表结果集过大,导致nestloop耗时过长,超过一小时未返回结果,因此性能优化的关键是消除nestloop,让join使用更高效的hashjoin。从语义等价的角度消除any-clause,SQL改写如下:

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selectls_pid_cusr1,COALESCE(max(round(ym/365)),0)from(         (                   SELECT                             ls_pid_cusr1,(current_date-bthdate) as ym                   FROM calc_empfyc_c1_result_tmp_t1 t1,p10_md_tmp_t2 t2                   WHERE t1.ls_pid_cusr1 = t2.id and t1.ls_pid_cusr1 != t2.id15         )         union all         (                   SELECT                             ls_pid_cusr1,(current_date-bthdate) as ym                   FROM calc_empfyc_c1_result_tmp_t1 t1,p10_md_tmp_t2 t2                   WHERE t1.ls_pid_cusr1 = id15         ))GROUP BY ls_pid_cusr1;

优化后的SQL查询由两个等值join的子查询构成,而每个子查询都可以走更适合此场景的hashjoin。优化后的执行计划如下

优化后,从超过1个小时未返回结果优化到7s返回结果。

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