备注:测试数据库版本为MySQL 8.0

这个blog我们来聊聊MySQL的with语句
对于逻辑复杂的sql，with可以大大减少临时表的数量，提升代码的可读性、可维护性

MySQL 8.0终于开始支持with语句了，对于复杂查询，可以不用写那么多的临时表了。

如需要scott用户下建表及录入数据语句，可参考:
scott建表及录入数据sql脚本

语句结构:

with subquery_name1 as (subquery_body1),
        subquery_name2 as (subquery_body2)
...
select * from subquery_name1 a, subquery_name2 b
where a.col = b.col
...

优势
– 代码模块化
– 代码可读性增强
– 相同查询唯一化

一.提升代码的可读性和可维护性

需求:求每个部门的平均工资,以及剔除薪资低于1000的实习人员之后的平均工资

-- 求每个部门的平均工资,以及剔除薪资低于1000的实习人员之后的平均工资
-- 主查询的from后面跟了2个临时表，程序可读性不佳
select d.deptno, tmp1.avg_sal avg_sal1, tmp2.avg_sal avg_sal2
  from dept d
  left join (select e1.deptno, round(avg(ifnull(e1.sal, 0)), 2) avg_sal
               from emp e1
              group by e1.deptno) tmp1
    on d.deptno = tmp1.deptno
  left join (select e1.deptno, round(avg(ifnull(e1.sal, 0)), 2) avg_sal
               from emp e1
              where e1.sal > 1000
              group by e1.deptno) tmp2
    on d.deptno = tmp2.deptno;
    
    
-- 求每个部门的平均工资,以及剔除薪资低于1000的实习人员之后的平均工资
-- 2个临时表的定时语句通过with封装成子查询了，程序可读性增强
with tmp1 as
 (select e1.deptno, round(avg(ifnull(e1.sal, 0)), 2) avg_sal
    from emp e1
   group by e1.deptno),
tmp2 as
 (select e1.deptno, round(avg(ifnull(e1.sal, 0)), 2) avg_sal
    from emp e1
   where e1.sal > 1000
   group by e1.deptno)
select d.deptno, tmp1.avg_sal avg_sal1, tmp2.avg_sal avg_sal2
  from dept d
  left join tmp1
    on d.deptno = tmp1.deptno
  left join tmp2
    on d.deptno = tmp2.deptno;

mysql> -- 求每个部门的平均工资,以及剔除薪资低于1000的实习人员之后的平均工资
mysql> -- 主查询的from后面跟了2个临时表，程序可读性不佳
mysql> select d.deptno, tmp1.avg_sal avg_sal1, tmp2.avg_sal avg_sal2
    ->   from dept d
    ->   left join (select e1.deptno, round(avg(ifnull(e1.sal, 0)), 2) avg_sal
    ->                from emp e1
    ->               group by e1.deptno) tmp1
    ->     on d.deptno = tmp1.deptno
    ->   left join (select e1.deptno, round(avg(ifnull(e1.sal, 0)), 2) avg_sal
    ->                from emp e1
    ->               where e1.sal > 1000
    ->               group by e1.deptno) tmp2
    ->     on d.deptno = tmp2.deptno;
+--------+----------+----------+
| deptno | avg_sal1 | avg_sal2 |
+--------+----------+----------+
|     10 |  2916.67 |  2916.67 |
|     20 |  2175.00 |  2518.75 |
|     30 |  1566.67 |  1690.00 |
|     40 |     NULL |     NULL |
+--------+----------+----------+
4 rows in set (0.00 sec)

mysql>
mysql>
mysql> -- 求每个部门的平均工资,以及剔除薪资低于1000的实习人员之后的平均工资
mysql> -- 2个临时表的定时语句通过with封装成子查询了，程序可读性增强
mysql> with tmp1 as
    ->  (select e1.deptno, round(avg(ifnull(e1.sal, 0)), 2) avg_sal
    ->     from emp e1
    ->    group by e1.deptno),
    -> tmp2 as
    ->  (select e1.deptno, round(avg(ifnull(e1.sal, 0)), 2) avg_sal
    ->     from emp e1
    ->    where e1.sal > 1000
    ->    group by e1.deptno)
    -> select d.deptno, tmp1.avg_sal avg_sal1, tmp2.avg_sal avg_sal2
    ->   from dept d
    ->   left join tmp1
    ->     on d.deptno = tmp1.deptno
    ->   left join tmp2
    ->     on d.deptno = tmp2.deptno;
+--------+----------+----------+
| deptno | avg_sal1 | avg_sal2 |
+--------+----------+----------+
|     10 |  2916.67 |  2916.67 |
|     20 |  2175.00 |  2518.75 |
|     30 |  1566.67 |  1690.00 |
|     40 |     NULL |     NULL |
+--------+----------+----------+
4 rows in set (0.00 sec)

mysql>

二.with递归

用with递归构造数列

-- 用with递归构造1-10的数据
with RECURSIVE c(n) as
 (select 1   union all select n + 1 from c where n < 10)
select n from c;

-- 用with递归构造1-10的数据
mysql> with RECURSIVE c(n) as
    ->  (select 1   union all select n + 1 from c where n < 10)
    -> select n from c;
+------+
| n    |
+------+
|    1 |
|    2 |
|    3 |
|    4 |
|    5 |
|    6 |
|    7 |
|    8 |
|    9 |
|   10 |
+------+
10 rows in set (0.00 sec)

用with递归构造级联关系

with RECURSIVE emp2(ename,empno,mgr,lvl)
  as
   (select ename, empno, mgr, 1 lvl from emp where mgr is null
    union all
    select emp.ename, emp.empno, emp.mgr, e2.lvl+1
      from emp, emp2 e2
     where emp.mgr = e2.empno
   )
select lvl,
      concat(repeat('**',lvl),ename) nm
  from emp2
 order by lvl,ename
;

mysql> with RECURSIVE emp2(ename,empno,mgr,lvl)
    ->   as
    ->    (select ename, empno, mgr, 1 lvl from emp where mgr is null
    ->     union all
    ->     select emp.ename, emp.empno, emp.mgr, e2.lvl+1
    ->       from emp, emp2 e2
    ->      where emp.mgr = e2.empno
    ->    )
    -> select lvl,
    ->       concat(repeat('**',lvl),ename) nm
    ->   from emp2
    ->  order by lvl,ename
    -> ;
+------+---------------+
| lvl  | nm            |
+------+---------------+
|    1 | **KING        |
|    2 | ****BLAKE     |
|    2 | ****CLARK     |
|    2 | ****JONES     |
|    3 | ******ALLEN   |
|    3 | ******FORD    |
|    3 | ******JAMES   |
|    3 | ******MARTIN  |
|    3 | ******MILLER  |
|    3 | ******SCOTT   |
|    3 | ******TURNER  |
|    3 | ******WARD    |
|    4 | ********ADAMS |
|    4 | ********SMITH |
+------+---------------+
14 rows in set (0.00 sec)