动力节点 mysql 郭鑫 34道经典的面试题
2024-09-03 22:40:43
DROP TABLE IF EXISTS `dept`;
CREATE TABLE `dept` (
`DEPTNO` int(2) NOT NULL COMMENT '部门编号',
`DNAME` varchar(14) DEFAULT NULL COMMENT '部门名称',
`LOC` varchar(13) DEFAULT NULL COMMENT '位置',
PRIMARY KEY (`DEPTNO`)
) ENGINE=InnoDB;
INSERT INTO `dept` VALUES ('', 'ACCOUNTING', 'NEW YORK');
INSERT INTO `dept` VALUES ('', 'RESEARCH', 'DALLAS');
INSERT INTO `dept` VALUES ('', 'SALES', 'CHICAGO');
INSERT INTO `dept` VALUES ('', 'OPERATIONS', 'BOSTON');
创建员工表
DROP TABLE IF EXISTS `emp`;
CREATE TABLE `emp` (
`EMPNO` int(4) NOT NULL COMMENT '员工编号',
`ENAME` varchar(10) DEFAULT NULL COMMENT '员工姓名',
`JOB` varchar(9) DEFAULT NULL COMMENT '工作岗位',
`MGR` int(4) DEFAULT NULL COMMENT '上级经理',
`HIREDATE` date DEFAULT NULL,
`SAL` double(7,2) DEFAULT NULL,
`COMM` double(7,2) DEFAULT NULL,
`DEPTNO` int(2) DEFAULT NULL,
PRIMARY KEY (`EMPNO`),
KEY `DEPTNO` (`DEPTNO`),
KEY `SAL` (`SAL`),
CONSTRAINT `emp_ibfk_1` FOREIGN KEY (`DEPTNO`) REFERENCES `dept` (`DEPTNO`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COMMENT='员工编号';
插入语句:
INSERT INTO `emp` VALUES ('', 'SMITH', 'CLERK', '', '1980-12-17', '800.00', null, '');
INSERT INTO `emp` VALUES ('', 'ALLEN', 'SALESMAN', '', '1981-02-20', '1600.00', '300.00', '');
INSERT INTO `emp` VALUES ('', 'WARD', 'SALESMAN', '', '1981-02-22', '1250.00', '500.00', '');
INSERT INTO `emp` VALUES ('', 'JONES', 'MANAGER', '', '1981-04-02', '2975.00', null, '');
INSERT INTO `emp` VALUES ('', 'MARTIN', 'SALESMAN', '', '1981-09-28', '1250.00', '1400.00', '');
INSERT INTO `emp` VALUES ('', 'BLAKE', 'MANAGER', '', '1981-05-01', '2850.00', null, '');
INSERT INTO `emp` VALUES ('', 'CLARK', 'MANAGER', '', '1981-06-09', '2450.00', null, '');
INSERT INTO `emp` VALUES ('', 'SCOTT', 'ANALYST', '', '1987-04-19', '3000.00', null, '');
INSERT INTO `emp` VALUES ('', 'KING', 'PRESIDENT', null, '1981-11-17', '5000.00', null, '');
INSERT INTO `emp` VALUES ('', 'TURNER', 'SALESMAN', '', '1981-09-08', '1500.00', '0.00', '');
INSERT INTO `emp` VALUES ('', 'ADAMS', 'CLERK', '', '1981-05-23', '1100.00', null, '');
INSERT INTO `emp` VALUES ('', 'JAMES', 'CLERK', '', '1981-12-03', '950.00', null, '');
INSERT INTO `emp` VALUES ('', 'FORD', 'ANALYST', '', '1981-12-03', '3000.00', null, '');
INSERT INTO `emp` VALUES ('', 'MILLER', 'CLERK', '', '1982-01-23', '1300.00', null, '');
C:薪水等级表
1,建表语句
DROP TABLE IF EXISTS `salgrade`;
CREATE TABLE `salgrade` (
`GRADE` int(11) DEFAULT NULL,
`LOSAL` int(11) DEFAULT NULL,
`HISAL` int(11) DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
INSERT INTO `salgrade` VALUES ('', '', '');
INSERT INTO `salgrade` VALUES ('', '', '');
INSERT INTO `salgrade` VALUES ('', '', '');
INSERT INTO `salgrade` VALUES ('', '', '');
INSERT INTO `salgrade` VALUES ('', '', '');
表的结构如下所示:
1.取得每个部门最高薪水的人员名称
第一步:求出每个部门的最高薪水
select
e.deptno,max(e.sal) as maxsal
from
emp e
group by
e.deptno;
+--------+---------+
| deptno | maxsal |
+--------+---------+
| 10 | 5000.00 |
| 20 | 3000.00 |
| 30 | 2850.00 |
+--------+---------+
将以上查询结果当成一个临时表t(deptno,maxsal)
select
e.deptno,e.ename,t.maxsal,e.sal
from
(select
e.deptno,max(e.sal) as maxsal
from
emp e
group by
e.deptno)t
join
emp e
on
t.deptno = e.deptno
where
t.maxsal = e.sal
order by
e.deptno;
+--------+-------+---------+---------+
| deptno | ename | maxsal | sal |
+--------+-------+---------+---------+
| 10 | KING | 5000.00 | 5000.00 |
| 20 | SCOTT | 3000.00 | 3000.00 |
| 20 | FORD | 3000.00 | 3000.00 |
| 30 | BLAKE | 2850.00 | 2850.00 |
+--------+-------+---------+---------+
分析下:
首先group by 首先经常和聚合函数max等配合使用,第二使用了group by 在select后面的查询字段只能是group by 后面指定的字段不能是其他字段
第三:join on 条件中 on 和where的却别,不清楚的看自己的博客
MYSQL LEFT JOIN操作中 ON与WHERE放置条件的区别
on是两个表联合查询连接起来生成一个临时表,where是在生成临时表的基础上,对生成的临时表进行条件帅选
t.deptno = e.deptno 表示两个表生成临时表的关系是 emp表中的部门编号必须等于 t表中的部门编号
where之后的条件是:emp表和t表已经生成了临时表,然后对临时表进行条件过滤 2.哪些人的薪水在部门平均薪水之上
2.哪些人的薪水在部门平均薪水之上
第一步:求出每个部门的平均薪水
select
e.deptno,avg(e.sal) as avgsal
from
emp e
group by
e.deptno;
+--------+-------------+
| deptno | avgsal |
+--------+-------------+
| 10 | 2916.666667 |
| 20 | 2175.000000 |
| 30 | 1566.666667 |
+--------+-------------+
将以上查询结果当成临时表t(deptno,avgsal) select
t.deptno,e.ename
from
(select
e.deptno,avg(e.sal) as avgsal
from
emp e
group by
e.deptno) t
join
emp e
on
e.deptno = t.deptno
where
e.sal > t.avgsal;
+--------+-------+
| deptno | ename |
+--------+-------+
| 30 | ALLEN |
| 20 | JONES |
| 30 | BLAKE |
| 20 | SCOTT |
| 10 | KING |
| 20 | FORD |
+--------+-------+
3.取得部门中(所有人的)平均薪水等级
第一种情况:emp表中按照部门进行分组,求出每个组的平均工资,看每个组的平均工资属于那个等级
第一步:求出部门的平均薪水
select
e.deptno,avg(e.sal) as avgsal
from
emp e
group by
e.deptno;
将以下查询结果当成临时表t(deptno,avgsal)
+--------+-------------+
| deptno | avgsal |
+--------+-------------+
| 10 | 2916.666667 |
| 20 | 2175.000000 |
| 30 | 1566.666667 |
+--------+-------------+ select * from salgrade;
+-------+-------+-------+
| GRADE | LOSAL | HISAL |
+-------+-------+-------+
| 1 | 700 | 1200 |
| 2 | 1201 | 1400 |
| 3 | 1401 | 2000 |
| 4 | 2001 | 3000 |
| 5 | 3001 | 9999 |
+-------+-------+-------+ select
t.deptno,t.avgsal,s.grade
from
(select
e.deptno,avg(e.sal) as avgsal
from
emp e
group by
e.deptno) t
join
salgrade s
on
t.avgsal between s.losal and s.hisal;
+--------+-------------+-------+
| deptno | avgsal | grade |
+--------+-------------+-------+
| 30 | 1566.666667 | 3 |
| 10 | 2916.666667 | 4 |
| 20 | 2175.000000 | 4 |
+--------+-------------+-------+
第二种情况:首先求出每个人的薪水属于那个等级,然后进行分组
3.2 取得部门中所有人的平均的薪水等级
第一步:求出每个人的薪水等级
select
e.deptno,e.ename,s.grade
from
emp e
join
salgrade s
on
e.sal between s.losal and s.hisal
order by
e.deptno;
+--------+--------+-------+
| deptno | ename | grade |
+--------+--------+-------+
| 10 | CLARK | 4 |
| 10 | MILLER | 2 |
| 10 | KING | 5 |
| 20 | ADAMS | 1 |
| 20 | SMITH | 1 |
| 20 | FORD | 4 |
| 20 | SCOTT | 4 |
| 20 | JONES | 4 |
| 30 | BLAKE | 4 |
| 30 | JAMES | 1 |
| 30 | ALLEN | 3 |
| 30 | WARD | 2 |
| 30 | TURNER | 3 |
| 30 | MARTIN | 2 |
+--------+--------+-------+
将以上查询结果当成临时表t(deptno,ename,grade)
select
t.deptno,avg(t.grade) as avgGrade
from
(select
e.deptno,e.ename,s.grade
from
emp e
join
salgrade s
on
e.sal between s.losal and s.hisal) t
group by
t.deptno;
+--------+----------+
| deptno | avgGrade |
+--------+----------+
| 10 | 3.6667 |
| 20 | 2.8000 |
| 30 | 2.5000 |
+--------+----------+
4.不准用组函数(MAX),取得最高薪水(给出两种解决方案)
select sal from emp order by sal desc limit 1;
5.取得平均薪水最高的部门的部门编号
5.取得平均薪水最高的部门的部门编号
第一步:求出部门平均薪水
select
e.deptno,avg(e.sal) as avgsal
from
emp e
group by
e.deptno;
+--------+-------------+
| deptno | avgsal |
+--------+-------------+
| 10 | 2916.666667 |
| 20 | 2175.000000 |
| 30 | 1566.666667 |
+--------+-------------+
第二步:将以上查询结果当成临时表t(deptno,avgsal),求出最高的平均薪水
select max(t.avgsal) as maxAvgSal from (select
e.deptno,avg(e.sal) as avgsal
from
emp e
group by
e.deptno) t;
+-------------+
| maxAvgSal |
+-------------+
| 2916.666667 |
+-------------+ select
e.deptno,avg(e.sal) as avgsal
from
emp e
group by
e.deptno
having
avgsal = (select max(t.avgsal) as maxAvgSal from (select
e.deptno,avg(e.sal) as avgsal
from
emp e
group by
e.deptno) t);
+--------+-------------+
| deptno | avgsal |
+--------+-------------+
| 10 | 2916.666667 |
+--------+-------------+
主要这里有一个坑不能写成下面的形式:
select
e.deptno,avg(e.sal) avgSal
from
emp e
group by
e.deptno
order by
avgSal desc
limit 1;
因为如果有100个部门,可能存在很多个部门的平均值都是一样的
6.取得平均薪水最高的部门的部门名称
select
e.deptno,d.dname,avg(e.sal) as avgsal
from
emp e
join
dept d
on
e.deptno = d.deptno
group by
e.deptno,d.dname
having
avgsal = (select max(t.avgsal) as maxAvgSal from (select
e.deptno,avg(e.sal) as avgsal
from
emp e
group by
e.deptno) t); 7.求平均薪水的等级最低的部门的部门名称
第一步:部门的平均薪水
select
e.deptno,d.dname,avg(e.sal) as avgsal
from
emp e
join
dept d
on
e.deptno = d.deptno
group by
e.deptno,d.dname;
+--------+------------+-------------+
| deptno | dname | avgsal |
+--------+------------+-------------+
| 10 | ACCOUNTING | 2916.666667 |
| 20 | RESEARCH | 2175.000000 |
| 30 | SALES | 1566.666667 |
+--------+------------+-------------+
第二步:将以上结果当成临时表t(deptno,avgsal)与salgrade表进行表连接:t.avgsal between s.losal and s.hisal;
select
t.deptno,t.dname,s.grade
from
(select
e.deptno,d.dname,avg(e.sal) as avgsal
from
emp e
join
dept d
on
e.deptno = d.deptno
group by
e.deptno,d.dname)t
join
salgrade s
on
t.avgsal between s.losal and s.hisal;
+--------+------------+-------+
| deptno | dname | grade |
+--------+------------+-------+
| 30 | SALES | 3 |
| 10 | ACCOUNTING | 4 |
| 20 | RESEARCH | 4 |
+--------+------------+-------+ 第三步:将以上查询结果当成一张临时表t
select min(t.grade) as minGrade from (select
t.deptno,t.dname,s.grade
from
(select
e.deptno,d.dname,avg(e.sal) as avgsal
from
emp e
join
dept d
on
e.deptno = d.deptno
group by
e.deptno,d.dname)t
join
salgrade s
on
t.avgsal between s.losal and s.hisal)t;
+----------+
| minGrade |
+----------+
| 3 |
+----------+ select
t.deptno,t.dname,s.grade
from
(select
e.deptno,d.dname,avg(e.sal) as avgsal
from
emp e
join
dept d
on
e.deptno = d.deptno
group by
e.deptno,d.dname)t
join
salgrade s
on
t.avgsal between s.losal and s.hisal
where
s.grade = (select min(t.grade) as minGrade from (select
t.deptno,t.dname,s.grade
from
(select
e.deptno,d.dname,avg(e.sal) as avgsal
from
emp e
join
dept d
on
e.deptno = d.deptno
group by
e.deptno,d.dname)t
join
salgrade s
on
t.avgsal between s.losal and s.hisal)t);
+--------+-------+-------+
| deptno | dname | grade |
+--------+-------+-------+
| 30 | SALES | 3 |
+--------+-------+-------+
8.取得比普通员工(员工代码没有在mgr上出现的)的最高薪水还要高的经理人姓名
第一步:找出普通员工(员工代码没有出现在mgr上的)
1.1 先找出mgr有哪些人
select distinct mgr from emp;
+------+
| mgr |
+------+
| 7902 |
| 7698 |
| 7839 |
| 7566 |
| NULL |
| 7788 |
| 7782 |
+------+
select * from emp where empno in(select distinct mgr from emp);
+-------+-------+-----------+------+------------+---------+------+--------+
| EMPNO | ENAME | JOB | MGR | HIREDATE | SAL | COMM | DEPTNO |
+-------+-------+-----------+------+------------+---------+------+--------+
| 7566 | JONES | MANAGER | 7839 | 1981-04-02 | 2975.00 | NULL | 20 |
| 7698 | BLAKE | MANAGER | 7839 | 1981-05-01 | 2850.00 | NULL | 30 |
| 7782 | CLARK | MANAGER | 7839 | 1981-06-09 | 2450.00 | NULL | 10 |
| 7788 | SCOTT | ANALYST | 7566 | 1987-04-19 | 3000.00 | NULL | 20 |
| 7839 | KING | PRESIDENT | NULL | 1981-11-17 | 5000.00 | NULL | 10 |
| 7902 | FORD | ANALYST | 7566 | 1981-12-03 | 3000.00 | NULL | 20 |
+-------+-------+-----------+------+------------+---------+------+--------+
select max(sal) as maxsal from emp where empno not in(select distinct mgr from emp where mgr is not null);
+---------+
| maxsal |
+---------+
| 1600.00 |
+---------+ not in不会自动忽略空值
in会自动忽略空值 select ename from emp where sal > (select max(sal) as maxsal from emp where empno not in(select distinct mgr from emp where mgr is not null));
+-------+
| ename |
+-------+
| JONES |
| BLAKE |
| CLARK |
| SCOTT |
| KING |
| FORD |
+-------+
第一步:找出普通员工(员工代码没有出现在mgr上的)
1.1 先找出mgr有哪些人
select distinct mgr from emp;
+------+
| mgr |
+------+
| 7902 |
| 7698 |
| 7839 |
| 7566 |
| NULL |
| 7788 |
| 7782 |
+------+
select * from emp where empno in(select distinct mgr from emp);
+-------+-------+-----------+------+------------+---------+------+--------+
| EMPNO | ENAME | JOB | MGR | HIREDATE | SAL | COMM | DEPTNO |
+-------+-------+-----------+------+------------+---------+------+--------+
| 7566 | JONES | MANAGER | 7839 | 1981-04-02 | 2975.00 | NULL | 20 |
| 7698 | BLAKE | MANAGER | 7839 | 1981-05-01 | 2850.00 | NULL | 30 |
| 7782 | CLARK | MANAGER | 7839 | 1981-06-09 | 2450.00 | NULL | 10 |
| 7788 | SCOTT | ANALYST | 7566 | 1987-04-19 | 3000.00 | NULL | 20 |
| 7839 | KING | PRESIDENT | NULL | 1981-11-17 | 5000.00 | NULL | 10 |
| 7902 | FORD | ANALYST | 7566 | 1981-12-03 | 3000.00 | NULL | 20 |
+-------+-------+-----------+------+------------+---------+------+--------+
select max(sal) as maxsal from emp where empno not in(select distinct mgr from emp where mgr is not null);
+---------+
| maxsal |
+---------+
| 1600.00 |
+---------+ not in不会自动忽略空值
in会自动忽略空值 select ename from emp where sal > (select max(sal) as maxsal from emp where empno not in(select distinct mgr from emp where mgr is not null));
+-------+
| ename |
+-------+
| JONES |
| BLAKE |
| CLARK |
| SCOTT |
| KING |
| FORD |
+-------+
这里有一个很关键的地方,not in 没有排除null值,如果存在null值和not in 做计算,得带的值就是null
mysql> select * from emp where empno not in(select distinct mgr from emp);
Empty set mysql>
select distinct mgr from emp 的结果存在null值
这里not in 没有去掉null值得到的结果就是空
9.取得薪水最高的前五名员工
select * from emp order by sal desc limit 0,5;
+-------+-------+-----------+------+------------+---------+------+--------+
| EMPNO | ENAME | JOB | MGR | HIREDATE | SAL | COMM | DEPTNO |
+-------+-------+-----------+------+------------+---------+------+--------+
| 7839 | KING | PRESIDENT | NULL | 1981-11-17 | 5000.00 | NULL | 10 |
| 7788 | SCOTT | ANALYST | 7566 | 1987-04-19 | 3000.00 | NULL | 20 |
| 7902 | FORD | ANALYST | 7566 | 1981-12-03 | 3000.00 | NULL | 20 |
| 7566 | JONES | MANAGER | 7839 | 1981-04-02 | 2975.00 | NULL | 20 |
| 7698 | BLAKE | MANAGER | 7839 | 1981-05-01 | 2850.00 | NULL | 30 |
+-------+-------+-----------+------+------------+---------+------+--------+ 10.取得薪水最高的第六到第十名员工 select * from emp order by sal desc limit 5,5;
+-------+--------+----------+------+------------+---------+---------+--------+
| EMPNO | ENAME | JOB | MGR | HIREDATE | SAL | COMM | DEPTNO |
+-------+--------+----------+------+------------+---------+---------+--------+
| 7782 | CLARK | MANAGER | 7839 | 1981-06-09 | 2450.00 | NULL | 10 |
| 7499 | ALLEN | SALESMAN | 7698 | 1981-02-20 | 1600.00 | 300.00 | 30 |
| 7844 | TURNER | SALESMAN | 7698 | 1981-09-08 | 1500.00 | 0.00 | 30 |
| 7934 | MILLER | CLERK | 7782 | 1982-01-23 | 1300.00 | NULL | 10 |
| 7654 | MARTIN | SALESMAN | 7698 | 1981-09-28 | 1250.00 | 1400.00 | 30 |
+-------+--------+----------+------+------------+---------+---------+--------+ 11.取得最后入职的5名员工
select * from emp order by hiredate desc limit 5;
+-------+--------+---------+------+------------+---------+------+--------+
| EMPNO | ENAME | JOB | MGR | HIREDATE | SAL | COMM | DEPTNO |
+-------+--------+---------+------+------------+---------+------+--------+
| 7876 | ADAMS | CLERK | 7788 | 1987-05-23 | 1100.00 | NULL | 20 |
| 7788 | SCOTT | ANALYST | 7566 | 1987-04-19 | 3000.00 | NULL | 20 |
| 7934 | MILLER | CLERK | 7782 | 1982-01-23 | 1300.00 | NULL | 10 |
| 7902 | FORD | ANALYST | 7566 | 1981-12-03 | 3000.00 | NULL | 20 |
| 7900 | JAMES | CLERK | 7698 | 1981-12-03 | 950.00 | NULL | 30 |
+-------+--------+---------+------+------------+---------+------+--------+
12.取得每个薪水等级有多少员工
第一步:查询出每个员工的薪水等级
select
e.ename,s.grade
from
emp e
join
salgrade s
on
e.sal between s.losal and s.hisal
order by
s.grade;
+--------+-------+
| ename | grade |
+--------+-------+
| JAMES | 1 |
| SMITH | 1 |
| ADAMS | 1 |
| MILLER | 2 |
| WARD | 2 |
| MARTIN | 2 |
| ALLEN | 3 |
| TURNER | 3 |
| BLAKE | 4 |
| FORD | 4 |
| CLARK | 4 |
| SCOTT | 4 |
| JONES | 4 |
| KING | 5 |
+--------+-------+ 将以上查询结果当成临时表t(ename,grade)
select
t.grade,count(t.ename) as totalEmp
from
(select
e.ename,s.grade
from
emp e
join
salgrade s
on
e.sal between s.losal and s.hisal) t
group by
t.grade;
+-------+----------+
| grade | totalEmp |
+-------+----------+
| 1 | 3 |
| 2 | 3 |
| 3 | 2 |
| 4 | 5 |
| 5 | 1 |
+-------+----------+
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