1、第十五题
15.列出受雇日期早于其直接上级的所有员工编号、姓名、部门名称
思路一：第一步将emp a看成员工表，将emp b 看成领导表，员工表的mgr字段应该等于领导表的主键字段
mysql> select e.empno, e.ename from emp e join emp b on e.mgr = b.empno where e.hiredate < b.hiredate; +-------+-------+ | empno | ename | +-------+-------+ | 7369 | SMITH | | 7499 | ALLEN | | 7521 | WARD | | 7566 | JONES | | 7698 | BLAKE | | 7782 | CLARK | | 7876 | ADAMS | +-------+-------+ 7 rows in set
第二步：显示上面员工的部门名称，将emp a员工表和dept d进行关联
mysql> select d.dname, e.empno, e.ename from emp e join emp b on e.mgr = b.empno join dept d on e.deptno = d.deptno where e.hiredate < b.hiredate; +------------+-------+-------+ | dname | empno | ename | +------------+-------+-------+ | ACCOUNTING | 7782 | CLARK | | RESEARCH | 7369 | SMITH | | RESEARCH | 7566 | JONES | | RESEARCH | 7876 | ADAMS | | SALES | 7499 | ALLEN | | SALES | 7521 | WARD | | SALES | 7698 | BLAKE | +------------+-------+-------+ 7 rows in set
下面两个题主要考察下左连接和右连接
14.列出所有员工及领导的名字
mysql> select e.ename, b.ename as leadername from emp e left join emp b on e.mgr = b.empno; +--------+------------+ | ename | leadername | +--------+------------+ | SMITH | FORD | | ALLEN | BLAKE | | WARD | BLAKE | | JONES | KING | | MARTIN | BLAKE | | BLAKE | KING | | CLARK | KING | | SCOTT | JONES | | KING | NULL | | TURNER | BLAKE | | ADAMS | SCOTT | | JAMES | BLAKE | | FORD | JONES | | MILLER | CLARK | +--------+------------+ 14 rows in set
下面考查右连接
16.列出部门名称和这些部门的员工信息，同时列出那些没有员工的部门
mysql> select d.dname, e.* from emp e right join dept d on e.deptno = d.deptno; +------------+-------+--------+-----------+------+------------+------+------+--------+ | dname | EMPNO | ENAME | JOB | MGR | HIREDATE | SAL | COMM | DEPTNO | +------------+-------+--------+-----------+------+------------+------+------+--------+ | ACCOUNTING | 7782 | CLARK | MANAGER | 7839 | 1981-06-09 | 2450 | NULL | 10 | | ACCOUNTING | 7839 | KING | PRESIDENT | NULL | 1981-11-17 | 5000 | NULL | 10 | | ACCOUNTING | 7934 | MILLER | CLERK | 7782 | 1982-01-23 | 1300 | NULL | 10 | | RESEARCH | 7369 | SMITH | CLERK | 7902 | 1980-12-17 | 800 | NULL | 20 | | RESEARCH | 7566 | JONES | MANAGER | 7839 | 1981-04-02 | 2975 | NULL | 20 | | RESEARCH | 7788 | SCOTT | ANALYST | 7566 | 1987-04-19 | 3000 | NULL | 20 | | RESEARCH | 7876 | ADAMS | CLERK | 7788 | 1981-05-23 | 1100 | NULL | 20 | | RESEARCH | 7902 | FORD | ANALYST | 7566 | 1981-12-03 | 3000 | NULL | 20 | | SALES | 7499 | ALLEN | SALESMAN | 7698 | 1981-02-20 | 1600 | 300 | 30 | | SALES | 7521 | WARD | SALESMAN | 7698 | 1981-02-22 | 1250 | 500 | 30 | | SALES | 7654 | MARTIN | SALESMAN | 7698 | 1981-09-28 | 1250 | 1400 | 30 | | SALES | 7698 | BLAKE | MANAGER | 7839 | 1981-05-01 | 2850 | NULL | 30 | | SALES | 7844 | TURNER | SALESMAN | 7698 | 1981-09-08 | 1500 | 0 | 30 | | SALES | 7900 | JAMES | CLERK | 7698 | 1981-12-03 | 950 | NULL | 30 | | OPERATIONS | NULL | NULL | NULL | NULL | NULL | NULL | NULL | NULL | +------------+-------+--------+-----------+------+------------+------+------+--------+ 15 rows in set
下面考查having的使用，如果使用了groupby对数据进行设置之后，还需要对数据结构进行限制需要使用having
17.列出至少有5个员工的所有部门
第一步：先求出每个部门有多少员工，将emp a和部门表 dept d表进行关联，条件是e.deptno=d.deptno
第二步：然后通过分组e.deptno,过来count(e.ename) >= 5
mysql> select e.deptno,count(e.ename) as totalEmp from emp e group by e.deptno having totalEmp >= 5; +--------+----------+ | deptno | totalEmp | +--------+----------+ | 20 | 5 | | 30 | 6 | +--------+----------+ 2 rows in set
这里比较关键：第一点 使用了group by 字段，select 后面的字段只能是group by后面的字段e.deptno和聚合函数对应的字段count(e.ename) as totalEmp
第二点：现在要对聚合函数的结果进行过滤，totalEmp字段不是数据库中的字段，不能使用where进行限制，只能使用having
接下来：考察的是子查询，子查询是在一个数据库表中
18.列出薪水比“SMITH”多的所有员工信息
第一步：首先求出是，smith的工资
第二步：然后求出工资高于simith的
mysql> select * from emp where sal > (select sal from emp where ename = 'SMITH'); +-------+--------+-----------+------+------------+------+------+--------+ | EMPNO | ENAME | JOB | MGR | HIREDATE | SAL | COMM | DEPTNO | +-------+--------+-----------+------+------------+------+------+--------+ | 7499 | ALLEN | SALESMAN | 7698 | 1981-02-20 | 1600 | 300 | 30 | | 7521 | WARD | SALESMAN | 7698 | 1981-02-22 | 1250 | 500 | 30 | | 7566 | JONES | MANAGER | 7839 | 1981-04-02 | 2975 | NULL | 20 | | 7654 | MARTIN | SALESMAN | 7698 | 1981-09-28 | 1250 | 1400 | 30 | | 7698 | BLAKE | MANAGER | 7839 | 1981-05-01 | 2850 | NULL | 30 | | 7782 | CLARK | MANAGER | 7839 | 1981-06-09 | 2450 | NULL | 10 | | 7788 | SCOTT | ANALYST | 7566 | 1987-04-19 | 3000 | NULL | 20 | | 7839 | KING | PRESIDENT | NULL | 1981-11-17 | 5000 | NULL | 10 | | 7844 | TURNER | SALESMAN | 7698 | 1981-09-08 | 1500 | 0 | 30 | | 7876 | ADAMS | CLERK | 7788 | 1981-05-23 | 1100 | NULL | 20 | | 7900 | JAMES | CLERK | 7698 | 1981-12-03 | 950 | NULL | 30 | | 7902 | FORD | ANALYST | 7566 | 1981-12-03 | 3000 | NULL | 20 | | 7934 | MILLER | CLERK | 7782 | 1982-01-23 | 1300 | NULL | 10 | +-------+--------+-----------+------+------------+------+------+--------+ 13 rows in set
子查询2
21.列出在部门“SALES”<销售部>工作的员工的姓名，假定不知道销售部门的部门编号
select deptno from dept where dname = 'SALES'; +--------+ | deptno | +--------+ | 30 | +--------+ select ename from emp where deptno = (select deptno from dept where dname = 'SALES'); +--------+ | ename | +--------+ | ALLEN | | WARD | | MARTIN | | BLAKE | | TURNER | | JAMES |
子查询三：in
24.列出薪金等于部门30中员工的薪金的其它员工的姓名和薪金 select distinct sal from emp where deptno = 30; +---------+ | sal | +---------+ | 1600.00 | | 1250.00 | | 2850.00 | | 1500.00 | | 950.00 | +---------+ select ename,sal from emp where sal in(select distinct sal from emp where deptno = 30) and deptno <> 30;
19.列出所有“CLERK”（办事员）的姓名及其部门名称，部门人数
这是一个比较综合的题目
1、第一步在emp a表中查询出那些人的job岗位是办事员
2、将emp a表和dept d表相关联就可以得到职位是办事员的emp对应的部门名称
3、查询出每个部门对应的员工总数
4、将第三步的查询结果作为一个临时表t与第二步的查询结果进行关联，关联条件是t.deptno = d.deptno
select d.deptno,d.dname,e.ename from emp e join dept d on e.deptno = d.deptno where e.job = 'CLERK'; t1 +--------+------------+--------+ | deptno | dname | ename | +--------+------------+--------+ | 10 | ACCOUNTING | MILLER | 3 | 20 | RESEARCH | SMITH | 5 | 20 | RESEARCH | ADAMS | 5 | 30 | SALES | JAMES | 6 +--------+------------+--------+ 求出每个部门的员工数量 select e.deptno,count(e.ename) as totalEmp from emp e group by e.deptno; t2 +--------+----------+ | deptno | totalEmp | +--------+----------+ | 10 | 3 | | 20 | 5 | | 30 | 6 | +--------+----------+ select t1.deptno,t1.dname,t1.ename,t2.totalEmp from (select d.deptno,d.dname,e.ename from emp e join dept d on e.deptno = d.deptno where e.job = 'CLERK') t1 join (select e.deptno,count(e.ename) as totalEmp from emp e group by e.deptno) t2 on t1.deptno = t2.deptno; +--------+------------+--------+----------+ | deptno | dname | ename | totalEmp | +--------+------------+--------+----------+ | 10 | ACCOUNTING | MILLER | 3 | | 20 | RESEARCH | SMITH | 5 | | 20 | RESEARCH | ADAMS | 5 | | 30 | SALES | JAMES | 6 | +--------+------------+--------+----------+
下面考查在select 后面两个聚合函数的事业
20.列出最低薪水大于1500的各种工作及从事此工作的全部雇员人数
第一步：求出每种工作岗位的最低薪水，并且最低薪水大于15000
第二步：在第一步的基础上求出雇员数量(count *)
第一步：先求出每种工作岗位的最低薪水 select e.job,min(e.sal) as minsal from emp e group by e.job; +-----------+---------+ | job | minsal | +-----------+---------+ | ANALYST | 3000.00 | | CLERK | 800.00 | | MANAGER | 2450.00 | | PRESIDENT | 5000.00 | | SALESMAN | 1250.00 | +-----------+---------+ select e.job,min(e.sal) as minsal,count(e.ename) as totalEmp from emp e group by e.job having minsal > 1500; +-----------+---------+ | job | minsal | +-----------+---------+ | ANALYST | 3000.00 | | MANAGER | 2450.00 | | PRESIDENT | 5000.00 | +-----------+---------+ +-----------+---------+----------+ | job | minsal | totalEmp | +-----------+---------+----------+ | ANALYST | 3000.00 | 2 | | MANAGER | 2450.00 | 3 | | PRESIDENT | 5000.00 | 1 | +-----------+---------+----------+
接下来是对上面知识点的全部的一个综合的复习
22.列出薪金高于公司平均薪金的所有员工，所在部门、上级领导、雇员的工资等级
相当的经典
第一步：求出薪金高于公司平均薪金的所有员工
第二步：把第一步的结果当成临时表t 将临时表t和部门表 dept d 和工资等级表salary s进行关联，求出员工所在的部门，雇员的工资等级等
关联的条件是t.deptno = d.deptno t.salary betweent s.lower and high;
第三步：求出第一步条件下的所有的上级领导，因为有的员工没有上级领导需要使用left join 左连接
第一步：求出公司的平均薪水 select avg(sal) as avgsal from emp; +-------------+ | avgsal | +-------------+ | 2073.214286 | +-------------+ select d.dname, e.ename, b.ename as leadername, s.grade from emp e join dept d on e.deptno = d.deptno left join emp b on e.mgr = b.empno join salgrade s on e.sal between s.losal and s.hisal where e.sal > (select avg(sal) as avgsal from emp); +------------+-------+------------+-------+ | dname | ename | leadername | grade | +------------+-------+------------+-------+ | RESEARCH | JONES | KING | 4 | | SALES | BLAKE | KING | 4 | | ACCOUNTING | CLARK | KING | 4 | | RESEARCH | SCOTT | JONES | 4 | | ACCOUNTING | KING | NULL | 5 | | RESEARCH | FORD | JONES | 4 | +------------+-------+------------+-------+
23.列出与“SCOTT”从事相同工作的所有员工及部门名称 查询出SCOTT的工作岗位 select job from emp where ename = 'SCOTT'; +---------+ | job | +---------+ | ANALYST | +---------+ select d.dname, e.* from emp e join dept d on e.deptno = d.deptno where e.job = (select job from emp where ename = 'SCOTT'); +----------+-------+-------+---------+------+------------+---------+------+--------+ | dname | EMPNO | ENAME | JOB | MGR | HIREDATE | SAL | COMM | DEPTNO | +----------+-------+-------+---------+------+------------+---------+------+--------+ | RESEARCH | 7788 | SCOTT | ANALYST | 7566 | 1987-04-19 | 3000.00 | NULL | 20 | | RESEARCH | 7902 | FORD | ANALYST | 7566 | 1981-12-03 | 3000.00 | NULL | 20 | +----------+-------+-------+---------+------+------------+---------+------+--------+
25.列出薪金高于在部门30工作的所有员工的薪金的员工姓名和薪金、部门名称 第一步：先找出部门30的最高薪水 select max(sal) as maxsal from emp where deptno = 30; +---------+ | maxsal | +---------+ | 2850.00 | +---------+ select d.dname, e.ename, e.sal from emp e join dept d on e.deptno = d.deptno where e.sal > (select max(sal) as maxsal from emp where deptno = 30); +------------+-------+---------+ | dname | ename | sal | +------------+-------+---------+ | ACCOUNTING | KING | 5000.00 | | RESEARCH | JONES | 2975.00 | | RESEARCH | SCOTT | 3000.00 | | RESEARCH | FORD | 3000.00 | +------------+-------+---------+
这个题很关键
26.列出在每个部门工作的员工数量、平均工资和平均服务期限
第一步：求出每个部门对应的所有员工，这里使用了右连接，保证显示所有的部门，但是有的部门不存在员工，但是也必须把所有的部门显示出来
第二步：在第一步的基础上求出所有员工的数量，这里因为有的部门员工是null，所有不能使用count(*)，count(*)统计包含null，应该使用count（e.ename）
第三步：求出员工的平均工资，因为有的部门员工不存在，所以对应的工作也是null，这里需要null值做处理
用法说明
1
IFNULL(expr1,expr2)
如果 expr1 不是 NULL，IFNULL() 返回 expr1，否则它返回 expr2。
IFNULL()返回一个数字或字符串值，取决于它被使用的上下文环境。
第四步：
mysql> SELECT TO_DAYS(950501);
+---------------------------------------------------------+
| TO_DAYS(950501) |
+---------------------------------------------------------+
| 728779 |
+---------------------------------------------------------+
1 row in set (0.00 sec)
我们来看下面的代码
ount(*)对行的数目进行计算，包含NULL
count(column)对特定的列的值具有的行数进行计算，不包含NULL值。
select count(*) from test2 ; select count(id) from test2 ; select count(name) from test2 ; select count(name) from test2 where name is null;
当运行结果我们可以得出：3,3,2,0，为什么呢？让我来解释一下。首先count(1)指的并不是计算1的个数，而是指表的第一个字段，如果第一个字段没有建立索引，他的效率是很低的；而且count(column name)默认查询的是指定字段非空的个数，如果你想查询数据的所有行数，恰巧指定字段又是一个可存在空库数据的字段，那么得到的数据就不会是期望的值。再来说一下count(),在上述的count(column name)查询方式中，如果指定的column为限制为非空，那么mysql会将上述表达式转化为count()来进行查询。所以如果想要查询数据大小，在mysql中建议使用count(*)来执行，含义明了，速度还快。
同理：
28.列出所有部门的详细信息和人数
这里因为用到了左连接，同上面的题目，因为使用到了右连接可能存在null值，统计人数的时候不能使用count(*),而要使用count(e.ename)字段的值
select d.deptno,d.dname,d.loc,count(e.ename) as totalEmp from emp e right join dept d on e.deptno = d.deptno group by d.deptno,d.dname,d.loc; +--------+------------+----------+----------+ | deptno | dname | loc | totalEmp | +--------+------------+----------+----------+ | 10 | ACCOUNTING | NEW YORK | 3 | | 20 | RESEARCH | DALLAS | 5 | | 30 | SALES | CHICAGO | 6 | | 40 | OPERATIONS | BOSTON | 0 | +--------+------------+----------+----------+
29.列出各种工作的最低工资及从事此工作的雇员姓名
select e.job,min(e.sal) as minsal from emp e group by e.job; +-----------+---------+ | job | minsal | +-----------+---------+ | ANALYST | 3000.00 | | CLERK | 800.00 | | MANAGER | 2450.00 | | PRESIDENT | 5000.00 | | SALESMAN | 1250.00 | +-----------+---------+ 将以上查询结果当成临时表t(job,minsal) select e.ename from emp e join (select e.job,min(e.sal) as minsal from emp e group by e.job) t on e.job = t.job where e.sal = t.minsal; +--------+ | ename | +--------+ | SMITH | | WARD | | MARTIN | | CLARK | | SCOTT | | KING | | FORD | +--------+