1：emp表中查询公司总共有几个部门
注意，会查询出来大量重复的，使用函数distinct
select distinct job from scott.emp;
2：查询公司工资在1000-3000之间的人有哪些
使用函数between ...and..
select * from scott.emp where sal between 3000 and 5000;
3：查询公司没有奖金的人
使用null 和“” 不一样
select * from scott.emp where comm is null;
4：查询公司员工职位是‘manager‘,‘clerk‘ 的人
select * from scott.emp where lower(job) in(‘manager‘,‘clerk‘);
查询不是这两个职位的人
select * from scott.emp where upper(job) not in(‘MANAGER‘,‘CLERK‘);
5：查询工资最高的人
查询每个部门工资最高的人？--分组查询
select deptno,max(sal) from scott.emp group by deptno ;
完整版
29：-每个部门的最高薪水是多少
select * from scott.emp where (deptno,sal) in (select deptno,max(sal) from scott.emp group by deptno);
7：-查询比20部门总人数多的部门
select deptno from scott.emp group by deptno having count(*) >(select count(*) from scott.emp where deptno=20);
13：薪水由高到低排序( 降序排列 )
select empno,ename,sal from scott.emp order by sal desc;
14-按入职时间排序 , 入职时间越早排在前面
select hiredate from scott.emp order by hiredate;
15按部门排序 , 同一部门按薪水由高到低排序
select deptno,sal from scott.emp order by deptno,sal desc;
16计算员工的薪水总和是多少？sum()
select sum(sal)+sum(comm) from scott.emp;
17计算最早和最晚的员工入职时间
select min(hiredate),max(hiredate) from scott.emp;
（查最早和最晚入职的人的信息）
select * from emp where hireDate in (select max(hireDate) from emp)
UNION
select * from emp where hireDate in (select min(hireDate) from emp)
18按部门计算每个部门的最高和最低薪水分别是多少
select deptno,max(sal),min(sal) from scott.emp group by deptno order by deptno;
19计算每个部门的 薪水总和 和 平均薪水？
select deptno,sum(sal),avg(sal) from scott.emp group by deptno;
20按职位分组 , 每个职位的最高、最低薪水和人数？
select job,max(sal),min(sal),count(*) from scott.emp group by job;
21平均薪水大于5000元的部门数据`
select deptno,avg(sal) from scott.emp group by deptno having avg(sal)>2000;
22薪水总和大于20000元的部门数据
select deptno,sum(sal) from scott.emp group by deptno having sum(sal)>10000;
23：哪些职位的人数超过2个人
select job,count(*) from scott.emp group by job having count(*)>2;
24：查询最高薪水的是谁？-查询最低薪水的员工
select ename from scott.emp where sal=(select max(sal) from scott.emp);
25：谁的工资比smith高
select ename from scott.emp where sal>(select sal from scott.emp where lower(ename)=‘smith‘);
26销售部门有哪些职位存在
select job from scott.emp where deptno=(select deptno from scott.dept where lower(dname)=‘sales‘);
--思考:
-- 谁的工资比smith高这道题目中如果存在两个角simth的人会怎么样?
--all:满足全部
--any:任意一个
insert into scott.emp(empno,ename,sal) values(9527,‘smith‘,888);
select ename from scott.emp where sal>all(select sal from scott.emp where lower(ename)=‘smith‘);
27-查询谁和smith是同一个部门的员工
select empno,ename,deptno from scott.emp where deptno=(select deptno from scott.emp where lower(ename)=‘smith‘);
28查询谁是员JAMES的下属
select empno,ename from scott.emp where mgr in(select mgr from scott.emp where lower(ename)=‘james‘);
30哪个部门的人数比部门20的人数多
select deptno,count(*) from scott.emp group by deptno having count(*)>
(select count(*) from scott.emp where deptno=20);
31：哪些员工的薪水比公司的平均薪水低？
select empno,ename from scott.emp where sal<(select avg(sal) from scott.emp);
--------------------------------------------华丽的分割线-----------------------------------------------------------------
Navicat MySQL Data Transfer 员工管理
Source Server : wode
Source Server Version : 50022
Source Host : localhost:3306
Source Database : j121
Target Server Type : MYSQL
Target Server Version : 50022
File Encoding : 65001
Date: 2016-05-09 10:33:00
*/
SET FOREIGN_KEY_CHECKS=0;
-- ----------------------------
-- Table structure for `dept`
-- ----------------------------
DROP TABLE IF EXISTS `dept`;
CREATE TABLE `dept` (
`deptNo` int(11) NOT NULL,
`dname` varchar(20) default NULL,
`loc` varchar(40) default NULL,
PRIMARY KEY (`deptNo`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
-- ----------------------------
-- Records of dept
-- ----------------------------
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‘);
-- ----------------------------
-- Table structure for `emp`
-- ----------------------------
DROP TABLE IF EXISTS `emp`;
CREATE TABLE `emp` (
`id` int(11) NOT NULL auto_increment,
`empno` int(11) default NULL,
`eName` varchar(20) default NULL,
`job` varchar(10) default NULL,
`mgr` int(11) default NULL,
`hireDate` date default NULL,
`sal` decimal(7,2) default NULL,
`comm` decimal(7,2) default NULL,
`deptNo` int(11) default NULL,
PRIMARY KEY (`id`),
KEY `FK_emp_deptNo` (`deptNo`),
CONSTRAINT `FK_emp_deptNo` FOREIGN KEY (`deptNo`) REFERENCES `dept` (`deptNo`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
-- ----------------------------
-- Records of emp
-- ----------------------------
INSERT INTO emp VALUES (‘1‘, ‘7369‘, ‘smith‘, ‘clerk‘, ‘7902‘, ‘1980-12-17‘, ‘800.00‘, null, ‘20‘);
INSERT INTO emp VALUES (‘2‘, ‘7499‘, ‘allen‘, ‘salesman‘, ‘7698‘, ‘1981-02-20‘, ‘1600.00‘, ‘300.00‘, ‘30‘);
INSERT INTO emp VALUES (‘3‘, ‘7521‘, ‘ward‘, ‘salesman‘, ‘7698‘, ‘1981-02-22‘, ‘1250.00‘, ‘500.00‘, ‘30‘);
INSERT INTO emp VALUES (‘4‘, ‘7566‘, ‘jones‘, ‘manager‘, ‘7839‘, ‘1981-04-02‘, ‘2975.00‘, null, ‘20‘);
INSERT INTO emp VALUES (‘5‘, ‘7654‘, ‘martin‘, ‘salesman‘, ‘7698‘, ‘1981-09-28‘, ‘1250.00‘, ‘1400.00‘, ‘30‘);
INSERT INTO emp VALUES (‘6‘, ‘7698‘, ‘blake‘, ‘manager‘, ‘7839‘, ‘1981-05-01‘, ‘2850.00‘, null, ‘30‘);
INSERT INTO emp VALUES (‘7‘, ‘7782‘, ‘clark‘, ‘manager‘, ‘7839‘, ‘1981-06-09‘, ‘2450.00‘, null, ‘10‘);
INSERT INTO emp VALUES (‘8‘, ‘7788‘, ‘scott‘, ‘analyst‘, ‘7566‘, ‘1987-07-31‘, ‘3000.00‘, null, ‘20‘);
INSERT INTO emp VALUES (‘9‘, ‘7839‘, ‘king‘, ‘president‘, null, ‘1981-11-17‘, ‘5000.00‘, null, ‘10‘);
INSERT INTO emp VALUES (‘10‘, ‘7844‘, ‘turner‘, ‘salesman‘, ‘7698‘, ‘1981-09-08‘, ‘1500.00‘, ‘0.00‘, ‘30‘);
INSERT INTO emp VALUES (‘11‘, ‘7876‘, ‘adams‘, ‘clerk‘, ‘7788‘, ‘1987-07-13‘, ‘1100.00‘, null, ‘20‘);
INSERT INTO emp VALUES (‘12‘, ‘7900‘, ‘james‘, ‘clerk‘, ‘7698‘, ‘1981-12-03‘, ‘950.00‘, null, ‘30‘);
INSERT INTO emp VALUES (‘13‘, ‘7902‘, ‘ford‘, ‘analyst‘, ‘7566‘, ‘1981-12-03‘, ‘3000.00‘, null, ‘20‘);
INSERT INTO emp VALUES (‘14‘, ‘7934‘, ‘miller‘, ‘clerk‘, ‘7782‘, ‘1982-02-23‘, ‘1300.00‘, null, ‘10‘);