MySQL高级查询习题

提示：

emp员工表(empno员工号/ename员工姓名/job工作/mgr上级编号/hiredate受雇日期/sal薪金/comm佣金/deptno所属部门编号)

dept部门表(deptno部门编号/dname部门名称/loc地点)

工资 ＝ 薪金 ＋ 佣金

首先先知道主外键的关系，从提示中不难看出 dept 部门表是主键，emp 员工表是外键

在Emp 表中完成如下练习：

– 创建部门表

CREATE TABLE dept(

teptno VARCHAR(20) PRIMARY KEY,

dname VARCHAR(20),

loc VARCHAR(20));1

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– 创建员工表

CREATE TABLE emp(empno VARCHAR(20) PRIMARY KEY,

ename VARCHAR(20),

job VARCHAR(20),

mar VARCHAR(20),

hiredate DATE,

sal DOUBLE,

comm DOUBLE,

deptno VARCHAR(20) REFERENCES dept(deptno));1

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– 给部门表添加数据

INSERT INTO dept(teptno,dname,loc) VALUES("1","研发部","六楼");

INSERT INTO dept(teptno,dname,loc) VALUES("2","测试部","五楼");

INSERT INTO dept(teptno,dname,loc) VALUES("3","销售部","四楼");

INSERT INTO dept(teptno,dname,loc) VALUES("4","推广部","三楼");

INSERT INTO dept(teptno,dname,loc) VALUES("5","生产部","二楼");

INSERT INTO dept(teptno,dname,loc) VALUES("6","咨询部","一楼");1

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– 给员工表添加数据

INSERT INTO emp VALUES("001","小团团","java开发工程师","111","2020-01-15",10000,4000,"1");

INSERT INTO emp VALUES("002","韩冰","妲己","111","2019-02-14",20000,2000,"2");

INSERT INTO emp VALUES("003","呆妹","java咨询师","111","2019-03-14",10000,1000,"3");

INSERT INTO emp VALUES("004","Uzi","乌老师","111","2017-05-14",50000,5000,"4");

INSERT INTO emp VALUES("005","letme","理论大师","111","2017-04-14",20000,8000,"5");

INSERT INTO emp VALUES("006","七七饿了","金牌讲师","111","2019-08-14",10000,2000,"6");

INSERT INTO emp VALUES("007","小橙子","金牌讲师","111","2020-07-14",20000,5000,"6");

INSERT INTO emp VALUES("008","小橘子","java开发工程师","111","2020-09-25",50000,5000,"1");

INSERT INTO emp VALUES("010","小兔子","老师","111","2020-10-01",30000,000,"6");

INSERT INTO emp VALUES("011","达摩","峡谷艺术家","111","2017-09-14",10000,2000,"3");1

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1、在emp表中查询出所有的记录。

-- 1、在emp表中查询出所有的记录。

SELECT * FROM emp;1

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2、在emp表中查询出薪水大于1500的记录，并且按照薪水的降序排列。

-- 2、在emp表中查询出薪水大于1500的记录，并且按照薪水的降序排列。

SELECT * FROM emp ORDER BY sal+comm DESC;1

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3、在emp表中查询出comm字段为空值的记录。

-- 3、在emp表中查询出comm字段为空值的记录。

-- 先添加一个comm为空的数据

INSERT INTO emp (empno,ename,job,mar,hiredate,sal,deptno)VALUES("011","猪八戒","峡谷艺术家","111","2015-12.12","9000","2");

SELECT * FROM emp WHERE comm IS NULL;1

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4、查询出emp表中含有几个部门的记录。(用DISTINCT去除重复记录)

-- 4、查询出emp表中含有几个部门的记录。(用DISTINCT去除重复记录)

SELECT COUNT(DISTINCT deptno) FROM emp;1

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5、在emp表中查询出部门编号为5或6的记录。(要求使用IN关键字)

-- 5、在emp表中查询出部门编号为5或6的记录(要求使用IN关键字)

SELECT* FROM emp WHERE deptno IN(5,6);1

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6、在emp表中查询出姓名的第二个字母为A的记录。

-- 6、在emp表中查询出姓名的第二个字母为A的记录。

-- 先添加一条第二个字母为A的数据

INSERT INTO emp VALUES("012","AAA","峡谷艺术家","111","2017-09-18",10000,6000,"3");

SELECT * FROM emp WHERE ename LIKE "_A%";1

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7、查询出emp表中总共有多少条记录。

-- 7、查询出emp表中总共有多少条记录。

SELECT COUNT(*) FROM emp;1

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8、查询emp表中出每个部门的部门代码、薪水之和、平均薪水。

-- 8、查询emp表中出每个部门的部门代码、薪水之和、平均薪水

SELECT AVG(sal+IFNULL(comm,0))AS '平均薪水',SUM(sal+IFNULL(comm,0))AS '薪水之和',deptno AS '部门编号' FROM emp GROUP BY deptno;1

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不忘初心，砥砺前行

文章来源: blog.csdn.net，作者：小小小芒果，版权归原作者所有，如需转载，请联系作者。

原文链接：blog.csdn.net/weixin_45686583/article/details/110550183