mysql练习题一

一表

A:部门表

1，建表语句

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 DEFAULT CHARSET=utf8 COMMENT='部门信息表';

2，数据插入语句

INSERT INTO `dept` VALUES ('10', 'ACCOUNTING', 'NEW YORK');
INSERT INTO `dept` VALUES ('20', 'RESEARCH', 'DALLAS');
INSERT INTO `dept` VALUES ('30', 'SALES', 'CHICAGO');
INSERT INTO `dept` VALUES ('40', 'OPERATIONS', 'BOSTON');

B:员工表

1，建表语句

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';

2，数据插入语句

INSERT INTO `emp` VALUES ('7369', 'SMITH', 'CLERK', '7902', '1980-12-17', '800.00', null, '20');
INSERT INTO `emp` VALUES ('7499', 'ALLEN', 'SALESMAN', '7698', '1981-02-20', '1600.00', '300.00', '30');
INSERT INTO `emp` VALUES ('7521', 'WARD', 'SALESMAN', '7698', '1981-02-22', '1250.00', '500.00', '30');
INSERT INTO `emp` VALUES ('7566', 'JONES', 'MANAGER', '7839', '1981-04-02', '2975.00', null, '20');
INSERT INTO `emp` VALUES ('7654', 'MARTIN', 'SALESMAN', '7698', '1981-09-28', '1250.00', '1400.00', '30');
INSERT INTO `emp` VALUES ('7698', 'BLAKE', 'MANAGER', '7839', '1981-05-01', '2850.00', null, '30');
INSERT INTO `emp` VALUES ('7782', 'CLARK', 'MANAGER', '7839', '1981-06-09', '2450.00', null, '10');
INSERT INTO `emp` VALUES ('7788', 'SCOTT', 'ANALYST', '7566', '1987-04-19', '3000.00', null, '20');
INSERT INTO `emp` VALUES ('7839', 'KING', 'PRESIDENT', null, '1981-11-17', '5000.00', null, '10');
INSERT INTO `emp` VALUES ('7844', 'TURNER', 'SALESMAN', '7698', '1981-09-08', '1500.00', '0.00', '30');
INSERT INTO `emp` VALUES ('7876', 'ADAMS', 'CLERK', '7788', '1981-05-23', '1100.00', null, '20');
INSERT INTO `emp` VALUES ('7900', 'JAMES', 'CLERK', '7698', '1981-12-03', '950.00', null, '30');
INSERT INTO `emp` VALUES ('7902', 'FORD', 'ANALYST', '7566', '1981-12-03', '3000.00', null, '20');
INSERT INTO `emp` VALUES ('7934', 'MILLER', 'CLERK', '7782', '1982-01-23', '1300.00', null, '10');

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;

2，数据插入语句

INSERT INTO `salgrade` VALUES ('1', '700', '1200');
INSERT INTO `salgrade` VALUES ('2', '1201', '1400');
INSERT INTO `salgrade` VALUES ('3', '1401', '2000');
INSERT INTO `salgrade` VALUES ('4', '2001', '3000');
INSERT INTO `salgrade` VALUES ('5', '3001', '9999');

二题目

1，取得每个部门中的最高薪水的人员名称。

第一步：取得每个部门的最高薪水。

mysql>  select
    ->          deptno, max(sal) maxsal
    ->  from
    ->          emp
    ->  group by
    ->          deptno;
+--------+---------+
| deptno | maxsal  |
+--------+---------+
|     10 | 5000.00 |
|     20 | 3000.00 |
|     30 | 2850.00 |
+--------+---------+

第二步：把上面获得的结果作为临时表t，然后与emp表进行内连接中的等值连接，连接条件为emp表的deptno,sal和t中的deptno,maxsal相等。

mysql> select
    ->          e.deptno, e.ename, e.sal, t.maxsal
    ->  from
    ->          emp e
    ->  join
    ->          (select deptno, max(sal) maxsal from emp group by deptno) t
    ->  on
    ->          e.deptno = t.deptno
    ->  and
    ->          e.sal = t.maxsal;
+--------+-------+---------+---------+
| deptno | ename | sal     | maxsal  |
+--------+-------+---------+---------+
|     10 | KING  | 5000.00 | 5000.00 |
|     20 | SCOTT | 3000.00 | 3000.00 |
|     20 | FORD  | 3000.00 | 3000.00 |
|     30 | BLAKE | 2850.00 | 2850.00 |
+--------+-------+---------+---------+

2，哪些人的薪水在部门的平均薪水之上。
第一步：取得每个部门的平均薪水。

mysql> select
    ->          deptno, avg(sal) as avgsal
    -> from
    ->          emp
    -> group by
    ->          deptno;
+--------+-------------+
| deptno | avgsal      |
+--------+-------------+
|     10 | 2916.666667 |
|     20 | 2175.000000 |
|     30 | 1566.666667 |
+--------+-------------+

第二步： 把上面获得的结果作为临时表t，与emp表进行内连接中的等值连接， 连接的条件为emp中的deptno和t中的deptno相等，并且emp中的sal大于t中的avgsal。

mysql> select
    ->          e.deptno, e.ename, e.sal, t.avgsal
    -> from
    ->          emp e
    -> join
    ->          (select deptno, avg(sal) as avgsal from emp group by deptno) t
    -> on
    ->          e.deptno = t.deptno
    -> and
    ->          e.sal > t.avgsal;
+--------+-------+---------+-------------+
| deptno | ename | sal     | avgsal      |
+--------+-------+---------+-------------+
|     10 | KING  | 5000.00 | 2916.666667 |
|     20 | JONES | 2975.00 | 2175.000000 |
|     20 | SCOTT | 3000.00 | 2175.000000 |
|     20 | FORD  | 3000.00 | 2175.000000 |
|     30 | ALLEN | 1600.00 | 1566.666667 |
|     30 | BLAKE | 2850.00 | 1566.666667 |
+--------+-------+---------+-------------+

3，3.1取得部门（所有人的）平均薪水的等级。

第一步：取得每个部门的平均薪水。

mysql> select
    ->          deptno, avg(sal) as avgsal
    -> from
    ->          emp
    -> group by
    ->          deptno;
+--------+-------------+
| deptno | avgsal    |
+--------+-------------+
|     10 | 2916.666667 |
|     20 | 2175.000000 |
|     30 | 1566.666667 |
+--------+-------------+

第二步：把上面的结果作为临时表t与salgrade表进行内连接中的非等值连接，条件为临时表中的avgsal在salgrade表中losal和hisal之间。

mysql> select
    ->          t.deptno, s.grade
    -> from
    ->          salgrade s
    -> join
    ->          (select deptno, avg(sal) as avgsal from emp group by deptno) t
    -> on
    ->          t.avgsal between s.losal and hisal;
+--------+-------+
| deptno | grade |
+--------+-------+
|     30 |     3 |
|     10 |     4 |
|     20 |     4 |
+--------+-------+

3.2取得每个部门中所有人的薪水等级的平均等级。

第一步：取得部门每个人的薪水等级。

mysql> select
    ->          e.deptno, e.ename, s.grade
    -> from
    ->          emp e
    -> join
    ->          salgrade s
    -> on
    ->          e.sal between s.losal and hisal;
+--------+--------+-------+
| deptno | ename  | grade |
+--------+--------+-------+
|     20 | SMITH  |     1 |
|     30 | JAMES  |     1 |
|     20 | ADAMS  |     1 |
|     30 | WARD   |     2 |
|     30 | MARTIN |     2 |
|     10 | MILLER |     2 |
|     30 | TURNER |     3 |
|     30 | ALLEN  |     3 |
|     20 | JONES  |     4 |
|     30 | BLAKE  |     4 |
|     10 | CLARK  |     4 |
|     20 | SCOTT  |     4 |
|     20 | FORD   |     4 |
|     10 | KING   |     5 |
+--------+--------+-------+

第二步：将上面的结果作为临时表t，将deptno作为分组字段，用avg对grade求平均值。

mysql> 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 hisal) t
    -> group by
    ->          t.deptno;
+--------+----------+
| deptno | avggrade |
+--------+----------+
|     10 |   3.6667 |
|     20 |   2.8000 |
|     30 |   2.5000 |
+--------+----------+

4，不准用分组函数（Max），取得最高薪水（给出两种解决方案）。

第一种方法：对薪水字段进行降序排序，然后取前一个。

mysql> select
    ->          sal
    -> from
    ->          emp
    -> order by
    ->          sal desc
    -> limit
    ->          1;
+---------+
| sal     |
+---------+
| 5000.00 |
+---------+

第二中方法：

第一步：取出每个员工的薪水。

mysql> select
    ->          sal
    -> from
    ->          emp;
+---------+
| sal     |
+---------+
|  800.00 |
|  950.00 |
| 1100.00 |
| 1250.00 |
| 1250.00 |
| 1300.00 |
| 1500.00 |
| 1600.00 |
| 2450.00 |
| 2850.00 |
| 2975.00 |
| 3000.00 |
| 3000.00 |
| 5000.00 |
+---------+

第二步：把上面的结果作为临时表t，再把这张临时表t看成两张表，然后作内连接中的非等值连接，条件为前表中的sal小于后表中的sal，这样作为最大值的5000就被过滤掉了；

mysql> select
    ->          distinct t.sal as sal
    -> from
    ->          (select sal from emp) t
    -> join
    ->          (select sal from emp) t2
    -> on
    ->          t.sal < t2.sal;
+---------+
| sal     |
+---------+
|  800.00 |
|  950.00 |
| 1100.00 |
| 1250.00 |
| 1300.00 |
| 1500.00 |
| 1600.00 |
| 2450.00 |
| 2850.00 |
| 2975.00 |
| 3000.00 |
+---------+

第三步：选择薪水不在上面表中的薪水即为最高薪水。

mysql> select
    ->          sal
    -> from
    ->          emp
    -> where
    ->          sal not in (select distinct t.sal as sal from (select sal from emp) t join (select sal from emp) t2 on t
.sal < t2.sal);
+---------+
| sal     |
+---------+
| 5000.00 |
+---------+