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：

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

或者

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：

SELECT 
ls_pid_cusr1,COALESCE(max(round((current_date-bthdate)/365)),0)
FROM calc_empfyc_c1_result_tmp_t1 t1,p10_md_tmp_t2 t2
WHERE 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，这种不等值的join操作必须走nestloop，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走更高效的hashjoin。从语义等价的角度消除any-clause，SQL改写如下：

select
ls_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。