MySQL?with語(yǔ)句講解
備注:測(cè)試數(shù)據(jù)庫(kù)版本為MySQL 8.0
今天我們來(lái)聊聊MySQL的with語(yǔ)句
對(duì)于邏輯復(fù)雜的sql,with可以大大減少臨時(shí)表的數(shù)量,提升代碼的可讀性、可維護(hù)性
MySQL 8.0終于開(kāi)始支持with語(yǔ)句了,對(duì)于復(fù)雜查詢,可以不用寫那么多的臨時(shí)表了。
如需要scott用戶下建表及錄入數(shù)據(jù)語(yǔ)句,可參考:
scott建表及錄入數(shù)據(jù)sql腳本
語(yǔ)句結(jié)構(gòu):
with subquery_name1 as (subquery_body1), subquery_name2 as (subquery_body2) ... select * from subquery_name1 a, subquery_name2 b where a.col = b.col ...
優(yōu)勢(shì)
- – 代碼模塊化
- – 代碼可讀性增強(qiáng)
- – 相同查詢唯一化
一.提升代碼的可讀性和可維護(hù)性
需求:求每個(gè)部門的平均工資,以及剔除薪資低于1000的實(shí)習(xí)人員之后的平均工資
-- 求每個(gè)部門的平均工資,以及剔除薪資低于1000的實(shí)習(xí)人員之后的平均工資 -- 主查詢的from后面跟了2個(gè)臨時(shí)表,程序可讀性不佳 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; -- 求每個(gè)部門的平均工資,以及剔除薪資低于1000的實(shí)習(xí)人員之后的平均工資 -- 2個(gè)臨時(shí)表的定時(shí)語(yǔ)句通過(guò)with封裝成子查詢了,程序可讀性增強(qiáng) 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> -- 求每個(gè)部門的平均工資,以及剔除薪資低于1000的實(shí)習(xí)人員之后的平均工資 mysql> -- 主查詢的from后面跟了2個(gè)臨時(shí)表,程序可讀性不佳 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> -- 求每個(gè)部門的平均工資,以及剔除薪資低于1000的實(shí)習(xí)人員之后的平均工資 mysql> -- 2個(gè)臨時(shí)表的定時(shí)語(yǔ)句通過(guò)with封裝成子查詢了,程序可讀性增強(qiáng) 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遞歸
用with遞歸構(gòu)造數(shù)列
-- 用with遞歸構(gòu)造1-10的數(shù)據(jù) with RECURSIVE c(n) as (select 1 union all select n + 1 from c where n < 10) select n from c;
-- 用with遞歸構(gòu)造1-10的數(shù)據(jù) 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遞歸構(gòu)造級(jí)聯(lián)關(guān)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)
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