动力节点 mysql 郭鑫 34道经典的面试题三

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   |
+--------+