MySQL作业题(补充:在mysql中计算两个日期的差值)

Posted Zephyr丶J

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mysql作业题

三个表情况

mysql> select * from emp; // 从emp表查询所有数据。* 代表所有
+-------+--------+-----------+------+------------+---------+---------+--------+
| EMPNO | ENAME  | JOB       | MGR  | HIREDATE   | SAL     | COMM    | DEPTNO |
+-------+--------+-----------+------+------------+---------+---------+--------+
|  7369 | SMITH  | CLERK     | 7902 | 1980-12-17 |  800.00 |    NULL |     20 |
|  7499 | ALLEN  | SALESMAN  | 7698 | 1981-02-20 | 1600.00 |  300.00 |     30 |
|  7521 | WARD   | SALESMAN  | 7698 | 1981-02-22 | 1250.00 |  500.00 |     30 |
|  7566 | JONES  | MANAGER   | 7839 | 1981-04-02 | 2975.00 |    NULL |     20 |
|  7654 | MARTIN | SALESMAN  | 7698 | 1981-09-28 | 1250.00 | 1400.00 |     30 |
|  7698 | BLAKE  | MANAGER   | 7839 | 1981-05-01 | 2850.00 |    NULL |     30 |
|  7782 | CLARK  | MANAGER   | 7839 | 1981-06-09 | 2450.00 |    NULL |     10 |
|  7788 | SCOTT  | ANALYST   | 7566 | 1987-04-19 | 3000.00 |    NULL |     20 |
|  7839 | KING   | PRESIDENT | NULL | 1981-11-17 | 5000.00 |    NULL |     10 |
|  7844 | TURNER | SALESMAN  | 7698 | 1981-09-08 | 1500.00 |    0.00 |     30 |
|  7876 | ADAMS  | CLERK     | 7788 | 1987-05-23 | 1100.00 |    NULL |     20 |
|  7900 | JAMES  | CLERK     | 7698 | 1981-12-03 |  950.00 |    NULL |     30 |
|  7902 | FORD   | ANALYST   | 7566 | 1981-12-03 | 3000.00 |    NULL |     20 |
|  7934 | MILLER | CLERK     | 7782 | 1982-01-23 | 1300.00 |    NULL |     10 |
+-------+--------+-----------+------+------------+---------+---------+--------+

mysql> select * from dept;
+--------+------------+----------+
| DEPTNO | DNAME      | LOC      |
+--------+------------+----------+
|     10 | ACCOUNTING | NEW YORK |
|     20 | RESEARCH   | DALLAS   |
|     30 | SALES      | CHICAGO  |
|     40 | OPERATIONS | BOSTON   |
+--------+------------+----------+

mysql> select * from salgrade;
+-------+-------+-------+
| GRADE | LOSAL | HISAL |
+-------+-------+-------+
|     1 |   700 |  1200 |
|     2 |  1201 |  1400 |
|     3 |  1401 |  2000 |
|     4 |  2001 |  3000 |
|     5 |  3001 |  9999 |
+-------+-------+-------+

1、取得每个部门最高薪水的人员名称

没想到第一个就写了老半天,害!
自己写的
先找每个部分最高薪水,形成一个表,并起别名为n(不起别名报错)
将员工表和表n连接
根据薪水相等,找出这四个人
在这里插入图片描述
看视频后:
我怎么没有表连接的on语句啊…答案应该是侥幸对了

mysql> select e.ename,d.* from emp e
    -> join (select deptno,max(sal) maxsal from emp group by deptno) d
    -> on e.deptno = d.deptno and e.sal = d.maxsal;
+-------+--------+---------+
| ename | deptno | maxsal  |
+-------+--------+---------+
| BLAKE |     30 | 2850.00 |
| SCOTT |     20 | 3000.00 |
| KING  |     10 | 5000.00 |
| FORD  |     20 | 3000.00 |
+-------+--------+---------+
4 rows in set (0.01 sec)

2.哪些人的薪水在部门的平均薪水之上

根据上一个题的思路,先找部门平均薪水,形成表n

mysql> select avg(sal),deptno
    -> from emp
    -> group by deptno;
+-------------+--------+
| avg(sal)    | deptno |
+-------------+--------+
| 2916.666667 |     10 |
| 2175.000000 |     20 |
| 1566.666667 |     30 |
+-------------+--------+
3 rows in set (0.02 sec)

然后在e表中,找到部门相同,但是薪水大于平均薪水的人,很快昂

mysql> select e.ename,e.sal
-> from emp e
-> join (select avg(sal) sal,deptno from emp group by deptno) n
-> on e.sal > n.sal and e.deptno = n.deptno;
+-------+---------+
| ename | sal     |
+-------+---------+
| ALLEN | 1600.00 |
| JONES | 2975.00 |
| BLAKE | 2850.00 |
| SCOTT | 3000.00 |
| KING  | 5000.00 |
| FORD  | 3000.00 |
+-------+---------+
6 rows in set (0.03 sec)

3、取得部门中(所有人的)平均的薪水等级

第一步,取得所有人的薪水等级

mysql> select e.ename,e.deptno,sg.grade
-> from emp e
-> join salgrade sg
-> on e.sal between sg.losal and sg.hisal;
+--------+--------+-------+
| ename  | deptno | grade |
+--------+--------+-------+
| SMITH  |     20 |     1 |
| ALLEN  |     30 |     3 |
| WARD   |     30 |     2 |
| JONES  |     20 |     4 |
| MARTIN |     30 |     2 |
| BLAKE  |     30 |     4 |
| CLARK  |     10 |     4 |
| SCOTT  |     20 |     4 |
| KING   |     10 |     5 |
| TURNER |     30 |     3 |
| ADAMS  |     20 |     1 |
| JAMES  |     30 |     1 |
| FORD   |     20 |     4 |
| MILLER |     10 |     2 |
+--------+--------+-------+
14 rows in set (0.00 sec)

第二步,将这个表中的等级按部门平均

mysql> select deptno, avg(grade)
    -> from (select e.ename, e.deptno, sg.grade
    -> from emp e
    -> join salgrade sg
    -> on e.sal between sg.losal and sg.hisal) as n
    -> group by
    -> deptno;
+--------+------------+
| deptno | avg(grade) |
+--------+------------+
|     10 |     3.6667 |
|     20 |     2.8000 |
|     30 |     2.5000 |
+--------+------------+
3 rows in set (0.00 sec)

看视频后:
在进行表连接后没有必要把它看成临时表,而是直接在后面进行分组就可以了

mysql> select e.deptno, avg(s.grade)
    -> from emp e
    -> join salgrade s
    -> on e.sal between s.losal and s.hisal
    -> group by deptno;
+--------+--------------+
| deptno | avg(s.grade) |
+--------+--------------+
|     10 |       3.6667 |
|     20 |       2.8000 |
|     30 |       2.5000 |
+--------+--------------+
3 rows in set (0.00 sec)

4、不准用组函数(Max),取得最高薪水(给出两种解决方案)

方法一:
第一反应,排序以后形成一个表,然后取第一行
或者升序,统计数量,然后找最后一个

mysql> select sal from emp order by sal desc
-> limit 0,1;
+---------+
| sal     |
+---------+
| 5000.00 |
+---------+
1 row in set (0.00 sec)

方法二:
想不出来,看答案
神之操作a.sal < b.sal,怎么取的呢,就是对于b中的一个值,将a中所有比它小的都得到,
这样最大值就不会显示在这个表中

mysql> select a.sal
    -> from emp a
    -> join emp b
    -> on a.sal < b.sal;
    (下面显示部分)
+---------+
| sal     |
+---------+		
|  800.00 |		//b=800时,没有值符合条件;b= 1600,共有6个值符合条件
| 1250.00 |
| 1250.00 |
| 1500.00 |
| 1100.00 |
|  950.00 |
| 1300.00 |
|  800.00 |	//此时,b = 1250,共有3个值符合条件
| 1100.00 |
|  950.00 |
|  800.00 |	//b = 2975....
| 1600.00 |
| 1250.00 |
| 1250.00 |
| 2850.00 |
| 2450.00 |
......
+---------+
89 rows in set (0.00 sec)

然后用dinsinct去重

mysql> select distinct a.sal
-> from emp a
-> join emp b
-> on a.sal < b.sal;
+---------+
| sal     |
+---------+
|  800.00 |
| 1250.00 |
| 1500.00 |
| 1100.00 |
|  950.00 |
| 1300.00 |
| 1600.00 |
| 2850.00 |
| 2450.00 |
| 2975.00 |
| 3000.00 |
+---------+
11 rows in set (0.00 sec)

最后找到不等于上表中的值

mysql> select sal from emp
    -> where sal not in (
    -> select distinct a.sal
    -> from emp a
    -> join emp b
    -> on a.sal < b.sal);
+---------+
| sal     |
+---------+
| 5000.00 |
+---------+
1 row in set (0.01 sec)

看了这个答案,发现上面自己写的几道题应该是写的麻烦了,后面写注意

5.取得平均薪水最高的部门的部门编号(至少给出两种解决方案)

第一步,每个部门的平均薪水

mysql> select deptno, avg(sal) from emp group by deptno;
+--------+-------------+
| deptno | avg(sal)    |
+--------+-------------+
|     10 | 2916.666667 |
|     20 | 2175.000000 |
|     30 | 1566.666667 |
+--------+-------------+
3 rows in set (0.00 sec)

第二步,取得最高薪水部分对应编号
用max或者排序

mysql> select deptno, max(avgsal)
-> from (select deptno, avg(sal) avgsal from emp group by deptno) new;
+--------+-------------+
| deptno | max(avgsal) |
+--------+-------------+
|     10 | 2916.666667 |
+--------+-------------+
1 row in set (0.00 sec)

注意:这样写对吗,看似输出的部门编号正确
在用select deptno, avg(sal) from emp group by deptno;生成一个表以后,在表中查询最大值,和部门编号并没有任何关系
因此输出的10只是因为在10在第一行而顺序输出的编号,实则并不正确
如果将最大值改成最小值以后,输出明显错误

mysql> select deptno, min(avgsal)
-> from (select deptno, avg(sal) avgsal from emp group by deptno) new;
+--------+-------------+
| deptno | min(avgsal) |
+--------+-------------+
|     10 | 1566.666667 |
+--------+-------------+
1 row in set (0.00 sec)

所以这里应该是取这个最大值,然后去表中匹配对应部门编号
mysql> select max(avgsal)
-> from (select deptno, avg(sal) avgsal from emp group by deptno) new;
然后再与根据部门分组的平均值相匹配

mysql> select deptno, avg(sal) as avgsal
-> from emp
-> group by
-> deptno
-> having avgsal = (select max(new.avgsal) from (select deptno, avg(sal) avgsal from emp group by deptno) new);
+--------+-------------+
| deptno | avgsal      |
+--------+-------------+
|     10 | 2916.666667 |
+--------+-------------+
1 row in set (0.00 sec)

或者(排序)

mysql> select deptno,avg(sal) avgsal
-> from emp
-> group by deptno
-> order by avgsal desc
-> limit 0,1;
+--------+-------------+
| deptno | avgsal      |
+--------+-------------+
|     10 | 2916.666667 |
+--------+-------------+
1 row in set (0.00 sec)

6.取得平均薪水最高的部门的部门名称

接上一题,上一题找的是部分编号,这题找名称,需要连接另一个表,然后查询

mysql> select d.dname
    -> from dept d
    -> join (select deptno, avg(sal) avgsal from emp group by deptno order by avgsal desc limit 0,1) as t
    -> on d.deptno = t.deptno;
+------------+
| dname      |
+------------+
| ACCOUNTING |
+------------+
1 row in set (0.00 sec)

感觉我怎么呆呆的啊,看着一样,就呆呆把上面的结果套了一下,唉
可以先将两个表连接,然后,再根据部门编号分组,求平均值,然后排序取第一个

mysql> select d.dname, avg(e.sal) as avgsal
-> from emp e
-> join dept d
-> on d.deptno = e.deptno
-> group by e.deptno
-> order by avgsal desc
-> limit 0,1;
+------------+-------------+
| dname      | avgsal      |
+------------+-------------+
| ACCOUNTING | 2916.666667 |
+------------+-------------+
1 row in set (0.01 sec)

求平均薪水的等级最低的部门的部门名称

先求平均薪水

mysql> select deptno, avg(sal)
-> from emp
-> group by deptno;
+--------+-------------+
| deptno | avg(sal)    |
+--------+-------------+
|     10 | 2916.666667 |
|     20 | 2175.000000 |
|     30 | 1566.666667 |
+--------+-------------+
3 rows in set (0.00 sec)

再求平均薪水的等级最低的部门

mysql> select n.deptno, n.avgsal,s.grade
-> from salgrade s
-> join (select deptno, avg(sal) avgsal from emp group by deptno) as n
-> on n.avgsal between losal and hisal
-> order by grade asc
-> limit 0,1
-> ;
+--------+-------------+-------+
| deptno | avgsal      | grade |
+--------+-------------+-------+
|     30 | 1566.666667 |     3 |
+--------+-------------+-------+
1 row in set (0.00 sec)

再连接部门表得到部门名称

mysql> select d.dname
    -> from dept d
    -> join (
    -> select n.deptno, n.avgsal,s.grade
    -> from salgrade s
    -> join (select deptno, avg(sal) avgsal from emp group by deptno) as n
    -> on n.avgsal between losal and hisal
    -> order by grade asc
    -> limit 0,1) as t
    -> on d.deptno = t.deptno;
+-------+
| dname |
+-------+
| SALES |
+-------+
1 row in set (0.00 sec)

感觉还是繁琐了,后面看视频在学习咋改进吧,先自己都做完

看视频后,发现不能用limit取等级最低,因为等级最低的可能有多个部门,所以要先求出最低的等级
因为最低薪水的等级肯定是最低的,根据这个取最低的等级

最低平均薪水

mysql> select avg(sal) avgsal from emp group by deptno order by avgsal asc limit 1
    -> ;
+-------------+
| avgsal      |
+-------------+
| 1566.666667 |
+-------------+
1 row in set (0.00 sec)

最低平均薪水的等级

mysql> select grade from salgrade where (select avg(sal) avgsal from emp group by deptno order by avgsal asc limit 1) between losal and hisal;
+-------+
| grade |
+-------+
|     3 |
+-------+
1 row in set (0.00 sec)

平均薪水等级的部门名称

mysql> select t.*, s.grade
    -> from (select d.dname, avg(sal) as avgsal from emp e  join dept d on e.deptno = d.deptno group by d.dname) as t
    -> join salgrade s
    -> on t.avgsal between s.losal and s.hisal;
+------------+-------------+-------+
| dname      | avgsal      | grade |
+------------+-------------+-------+
| SALES      | 1566.666667 |     3 |
| ACCOUNTING | 2916.666667 |     4 |
| RESEARCH   | 2175.000000 |     4 |
+------------+-------------+-------+
3 rows in set (0.00 sec)

最低平均薪水等级的部门名称(条件:上表中的等级 等于 最低平局薪水等级)

mysql> select t.*, s.grade
    -> from (select d.dname, avg(sal) as avgsal from emp e  join dept d on e.deptno = d.deptno group by d.dname) as t
    -> join salgrade s
    -> on t.avgsal between s.losal and s.hisal
    -> where s.grade =
    -> (select grade from salgrade where (select avg(sal) avgsal from emp group by deptno order by avgsal asc limit 1) between losal and hisal);
+-------+-------------+-------+
| dname | avgsal      | grade |
+-------+-------------+-------+
| SALES | 1566.666667 |     3 |
+-------+-------------+-------+
1 row in set (0.00 sec)

8、取得比普通员工(员工代码没有在 mgr 字段上出现的)的最高薪水还要高的领导人姓名

这个题的是什么意思呢,就是不当领导的员工(普通员工)的最高薪水,然后求比这个最高薪水还高的领导人的姓名
所以肯定是自连接
先求两个表,一个是普通员工的薪水表,另一个是领导的薪水表

首先试了一下,这样报错

mysql> select empeno,ename,sal
    -> from emp
    -> where empno not in mgr;
ERROR 1064 (42000): You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'mgr' at line 3

然后就规规矩矩先找出所有领导的编号

mysql> select distinct mgr
-> from emp
-> where mgr is not null;
±-----+
| mgr |
±-----+
| 7902 |
| 7698 |
| 7839 |
| 7566 |
| 7788 |
| 7782 |
±-----+
6 rows in set (0.00 sec)

然后找出所有普通员工的编号和薪水,并排序
注意:not in在使用的时候,后面小括号中记得排除NULL。

mysql> select ename,sal,empno
    -> from emp
    -> where empno not in(
    -> select distinct mgr
    -> from emp
    -> where mgr is not null)
    -> order by sal desc
    -> ;
+--------+---------+-------+
| ename  | sal     | empno |
+--------+---------+-------+
| ALLEN  | 1600.00 |  7499 |
| TURNER | 1500.00 |  7844 |
| MILLER | 1300.00 |  7934 |
| WARD   | 1250.00 |  7521 |
| MARTIN | 1250.00 |  7654 |
| ADAMS  | 1100.00 |  7876 |
| JAMES  |  950.00 |  7900 |
| SMITH  |  800.00 |  7369 |
+--------+---------+-------+
8 rows in set (0.00 sec)

这时已经找到了普通员工的最高薪水,然后就找比这个薪水还高的领导姓名
找所有领导

mysql> select ename, sal, empno
    -> from emp
    -> where empno in(
    -> select distinct mgr
    -> from emp
    -> where mgr is not null);
+-------+---------+-------+
| ename | sal     | empno |
+-------+---------+-------+
| JONES | 2975.00 |  7566 |
| BLAKE | 2850.00 |  7698 |
| CLARK | 2450.00 |  7782 |
| SCOTT | 3000.00 |  7788 |
| KING  | 5000.00 |  7839 |
| FORD  | 3000.00 |  7902 |
+-------+---------+-------+
6 rows in set (0.00 sec)

找薪水大于最多薪水的领导姓名

mysql> select l.ename,l.sal
    -> from (select ename, sal from emp where empno in(select distinct mgr from emp where mgr is not null)) as l
    -> where l.sal > (select max(sal) from emp where empno not in(select distinct mgr from emp where mgr is not null));
+-------+---------+
| ename | sal     |
+-------+---------+
| JONES | 2975.00 |
| BLAKE | 2850.00 |
| CLARK | 2450.00 |
| SCOTT | 3000.00 |
| KING  | 5000.00 |
| FORD  | 3000.00 |
+-------+---------+
6 rows in set (0.00 sec)

这个题老师给的答案应该是错的,虽然答案是一样的
找的是领导人姓名,而不是在所有员工里找大于1600的

看视频后:
老师没错,我错了,比普通员工的最高薪水还高的一定是领导!!!!!
所以不用筛选领导,直接在所有员工里找就行了
select ename,sal from emp where sal > (select max(sal) from emp where empno not in(select distinct mgr from emp where mgr is not null));

9、取得薪水最高的前五名员工

突然简单,不太适应

mysql> select ename, sal
    -> from emp
    -> order by sal desc
    -> limit 0,5;
+-------+---------+
| ename | sal     |
+-------+---------+
| KING  | 5000.00 |
| SCOTT | 3000.00 |
| FORD  | 3000.00 |
| JONES | 2975.00 |
| BLAKE | 2850.00 |
+-------+---------+
5 rows in set (0.00 sec)

10、取得薪水最高的第六到第十名员工

mysql> select ename, sal
-> from emp
-> order by sal desc
-> limit 5,5;
+--------+---------+
| ename  | sal     |
+--------+---------+
| CLARK  | 2450.00 |
| ALLEN  | 1600.00 |
| TURNER | 1500.00 |
| MILLER | 1300.00 |
| MARTIN | 1250.00 |
+--------+---------+
5 rows in set (0.00 sec)

11、取得最后入职的 5 名员工

这个题给的知识点,可以按日期排序,并且从早到晚是升序,从晚到早是降序

mysql> select ename, hiredate
    -> from emp
    -> order by hiredate desc
    -> limit 0,5;
+--------+------------+
| ename  | hiredate   |
+--------+------------+
| ADAMS  | 1987-05-23 |
| SCOTT  | 1987-04-19 |
| MILLER | 1982-01-23 |
| FORD   | 1981-12-03 |
| JAMES  | 1981-12-03 |
+--------+------------+
5 rows in set (0.00 sec)

12、取得每个薪水等级有多少员工

先算薪水等级,然后计数

mysql> select e.ename, s.grade
    -> from emp e
    -> join salgrade s
    -> on e.sal between losal and hisal;
+--------+-------+
| ename  | grade |
+--------+-------+
| SMITH  |     1 |
| ALLEN  |     3 |
| WARD   |     2 |
| JONES  |     4 |
| MARTIN |     2 |
| BLAKE  |     4 |
| CLARK  |     4 |
| SCOTT  |     4 |
| KING   |     5 |
| TURNER |     3 |
| ADAMS  |     1 |
| JAMES  |     1 |
| FORD   |     4 |
| MILLER |     2 |
+--------+-------+
14 rows in set (0.00 sec)

计数

mysql> select grade, count(*)
    -> from (
    -> select e.ename, s.grade
    -> from emp e
    -> join salgrade s
    -> on e.sal between losal and hisal) as n
    -> group by n.grade;
+-------+----------+
| grade | count(*) |
+-------+----------+
|     1 |        3 |
|     2 |        3 |
|     3 |        2 |
|     4 |        5 |
|     5 |        1 |
+-------+----------+
5 rows in set (0.00 sec)

看视频后:
这里同样没必要把第一步的查询结果变成一个张表,直接分组查就行了
mysql> select s.grade,count(*)
-> from emp e
-> join salgrade s
-> on e.sal between losal and hisal
-> group by s.grade;

13、面试题

问题 1.找出没选过“黎明”老师的所有学生姓名。

先找出黎明老师的编号,然后再选课表中,查看选了黎明老师的学生的编号,然后在学生表中查看没有选黎明老师的学生姓名

先找出黎明老师的编号

mysql> select cno from c where cteacher = 'liming';
+------+
| cno  |
+------+
| 5    |
+------+
1 row in set (0.08 sec)

然后再选课表中,查看选了黎明老师的学生的编号,

mysql> select sno from sc
    -> where cno = (select cno from c where cteacher = 'liming');
+------+
| sno  |
+------+
| 1    |
| 2    |
+------+
2 rows in set (0.03 sec)

在学生表中查看没有选黎明老师的学生姓名

mysql> select sname from s
    -> where sno not in (
    -> select sno from sc
    -> where cno = (select cno from c where cteacher = 'liming'));
+-----------+
| sname     |
+-----------+
| xuesheng3 |
| xuesheng4 |
+-----------+
2 rows in set (0.03 sec)
问题 2.列出 2 门以上(含 2 门)不及格学生姓名及平均成绩。

先找出不及格的所有人和课程

mysql> select sno,cno,scgrade
    -> from sc where scgrade < 60;
+------+------+---------+
| sno  | cno  | scgrade |
+------+------+---------+
| 1    | 1    | 40      |
| 1    | 2    | 30      |
| 1    | 3    | 20      |
| 2    | 5    | 40      |
+------+------+---------+
4 rows in set (0.01 sec)

然后统计每个人不及格的课程数目,选出2门以上的学生编号

mysql> select sno,count(sno) from (select sno,cno,scgrade from sc where scgrade < 60) as n
    -> group by sno having count(sno) >= 2;
+------+------------+
| sno  | count(sno) |
+------+------------+
| 1    |          3 |
+------+------------+
1 row in set (0.12 sec)

找选课表中计算该学生的平均成绩

mysql> select sno,avg(scgrade) from sc where sno = (
-> select sno from (select sno,cno,scgrade from sc where scgrade < 60) as n
-> group by sno having count(sno) >= 2)
-> group by sno;
+------+--------------+
| sno  | avg(scgrade) |
+------+--------------+
| 1    |           46 |
+------+--------------+
1 row in set (0.07 sec)

在学生表中找学生名字,将两个表连接,根据学生编号

mysql> select a.sname, b.avggrade
    -> from s as a
    -> join (
    -> select sno,avg(scgrade) avggrade from sc where sno = (
    -> select sno from (select sno,cno,scgrade from sc where scgrade < 60) as n
    -> group by sno having count(sno) >= 2)
    -> group by sno) as b
    -> on a.sno = b.sno;
+-----------+----------+
| sname     | avggrade |
+-----------+----------+
| xuesheng1 |       46 |
+-----------+----------+
1 row in set (0.01 sec)
问题 3:既学过 1 号课程又学过 2 号课所有学生的姓名。

学过1号课程的学生,学过2号课程的学生

mysql> select sno,cno from sc where cno = '1';
+------+------+
| sno  | cno  |
+------+------+
| 1    | 1    |
| 2    | 1    |
| 3    | 1    |
+------+------+
3 rows in set (0.00 sec)

mysql> select sno,cno from sc where cno = '2';
+------+------+
| sno  | cno  |
+------+------+
| 1    | 2    |
| 2    | 2    |
+------+------+
2 rows in set (0.00 sec)

同时学过两门课程的学生

mysql> select a.sno from (select sno,cno from sc where cno = '1') as a
    -> where a.sno in (select b.sno from
    -> (select sno,cno from sc where cno = '2') as b);
+------+
| sno  |
+------+
| 1    |
| 2    |
+------+
2 rows in set (0.00 sec)

在学生表中找出姓名

mysql> select s.sname from s
    -> where sno in (
    -> select a.sno from (select sno,cno from sc where cno = '1') as a
    -> where a.sno in (select b.sno from
    -> (select sno,cno from sc where cno = '2') as b));
+-----------+
| sname     |
+-----------+
| xuesheng1 |
| xuesheng2 |
+-----------+
2 rows in set (0.00 sec)

14、列出所有员工及领导的姓名

相当于两张表,一张所有员工的表,一张关系表,然后通过关系表查对应的领导,注意因为boss没有领导,所以使用外连接,为NULL时处理

mysql> select a.ename, ifnull(b.ename, 'I\\'m boss') from emp a
    -> left join emp b
    -> on a.mgr = b.empno;
+--------+------------------------------+
| ename  | ifnull(b.ename, 'I\\'m boss') |
+--------+------------------------------+
| SMITH  | FORD                         |
| ALLEN  | BLAKE                        |
| WARD   | BLAKE                        |
| JONES  | KING                         |
| MARTIN | BLAKE                        |
| BLAKE  | KING                         |
| CLARK  | KING                         |
| SCOTT  | JONES                        |
| KING   | I'm boss                     |
| TURNER | BLAKE                        |
| ADAMS  | SCOTT                        |
| JAMES  | BLAKE                        |
| FORD   | JONES                        |
| MILLER | CLARK                        |
+--------+------------------------------+
14 rows in set (0.00 sec)

15、列出受雇日期早于其直接上级的所有员工的编号,姓名,部门名称

上一题找到了领导,这里判断它的受雇日期是否早于其领导

mysql> select a.empno, a.ename, b.ename,a.deptno
    -> from emp a
    -> join emp b
    -> on a.mgr = b.empno
    -> where a.hiredate < b.hiredate;
+-------+-------+-------+--------+
| empno | ename | ename | deptno |
+-------+-------+-------+--------+
|  7369 | SMITH | FORD  |     20 |
|  7499 | ALLEN | BLAKE |     30 |
|  7521 | WARD  | BLAKE |     30 |
|  7566 | JONES | KING  |     20 |
|  7698 | BLAKE | KING  |     30 |
|  7782 | CLARK | KING  |     10 |
+-------+-------+-------+--------+
6 rows in set (0.02 sec)

然后连接部门表,查找部门名称

mysql> select t.empno, t.ename, d.dname
    -> from (
    -> select a.empno, a.ename, a.deptno
    -> from emp a
    -> join emp b
    -> on a.mgr = b.empno
    -> where a.hiredate < b.hiredate) as t
    -> join dept d
    -> on t.deptno = d.deptno;
+-------+-------+------------+
| empno | ename | dname      |
+-------+-------+------------+
|  7782 | CLARK | ACCOUNTING |
|  7369 | SMITH | RESEARCH   |
|  7566 | JONES | RESEARCH   |
|  7499 | ALLEN | SALES      |
|  7521 | WARD  | SALES      |
|  7698 | BLAKE | SALES      |
+-------+-------+------------+

看视频后:
还是同样的问题,不需要嵌套,又繁琐了,直接在后面再连接就行

mysql> select a.ename, b.ename,a.deptno,d.dname
-> from emp a
-> join emp b
-> on a.mgr = b.empno
-> join dept d
-> a.deptno = d.deptno
-> where a.hiredate < b.hiredate;

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

左外链接

mysql> select d.dname,e.* from dept d left join emp e on d.deptno = e.deptno;
+------------+-------+--------+-----------+------+------------+---------+---------+--------+
| dname      | EMPNO | ENAME  | JOB       | MGR  | HIREDATE   | SAL     | COMM    | DEPTNO |
+------------+-------+--------+-----------+------+------------+---------+---------+--------+
| ACCOUNTING |  7782 | CLARK  | MANAGER   | 7839 | 1981-06-09 | 2450.00 |    NULL |     10 |
| ACCOUNTING |  7839 | KING   | PRESIDENT | NULL | 1981-11-17 | 5000.00 |    NULL |     10 |
| ACCOUNTING |  7934 | MILLER | CLERK     | 7782 | 1982-01-23 | 1300.00 |    NULL |     10 |
| RESEARCH   |  7369 | SMITH  | CLERK     | 7902 | 1980-12-17 |  800.00 |    NULL |     20 |
| RESEARCH   |  7566 | JONES  | MANAGER   | 7839 | 1981-04-02 | 2975.00 |    NULL |     20 |
| RESEARCH   |  7788 | SCOTT  | ANALYST   | 7566 | 1987-04-19 | 3000.00 |    NULL |     20 |
| RESEARCH   |  7876 | ADAMS  | CLERK     | 7788 | 1987-05-23 | 1100.00 |    NULL |     20 |
| RESEARCH   |  7902 | FORD   | ANALYST   | 7566 | 1981-12-03 | 3000.00 |    NULL |     20 |
| SALES      |  7499 | ALLEN  | SALESMAN  | 7698 | 1981-02-20 | 1600.00 |  300.00 |     30 |
| SALES      |  7521 | WARD   | SALESMAN  | 7698 | 1981-02-22 | 1250.00 |  500.00 |     30 |
| SALES      |  7654 | MARTIN | SALESMAN  | 7698 | 1981-09-28 | 1250.00 | 1400.00 |     30 |
| SALES      |  7698 | BLAKE  | MANAGER   | 7839 | 1981-05-01 | 2850.00 |    NULL |     30 |
| SALES      |  7844 | TURNER | SALESMAN  | 7698 | 1981-09-08 | 1500.00 |    0.00 |     30 |
| SALES      |  7900 | JAMES  | CLERK     | 7698 | 1981-12-03 |  950.00 |    NULL |     30 |
| OPERATIONS |  NULL | NULL   | NULL      | NULL | NULL       |    NULL |    NULL |   NULL |
+------------+-------+--------+-----------+------+------------+---------+---------+--------+
15 rows in set (0.00 sec)

17、列出至少有 5 个员工的所有部门

接上一题,分组统计完事

mysql> select d.dname,count(dname) from dept d left join emp e on d.deptno = e.deptno
    -> group by dname
    -> having count(dname) >= 5;
+----------+--------------+
| dname    | count(dname) |
+----------+--------------+
| RESEARCH |            5 |
| SALES    |            6 |
+----------+--------------+
2 rows in set (0.00 sec)

18、列出薪金比"SMITH"多的所有员工信息

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 | 1600.00 |  300.00 |     30 |
|  7521 | WARD   | SALESMAN  | 7698 | 1981-02-22 | 1250.00 |  500.00 |     30 |
|  7566 | JONES  | MANAGER   | 7839 | 1981-04-02 | 2975.00 |    NULL |     20 |
|  7654 | MARTIN | SALESMAN  | 7698 | 1981-09-28 | 1250.00 | 1400.00 |     30 |
|  7698 | BLAKE  | MANAGER   | 7839 | 1981-05-01 | 2850.00 |    NULL |     30 |
|  7782 | CLARK  | MANAGER   | 7839 | 1981-06-09 | 2450.00 |    NULL |     10 |
|  7788 | SCOTT  | ANALYST   | 7566 | 1987-04-19 | 3000.00 |    NULL |     20 |
|  7839 | KING   | PRESIDENT | NULL | 1981-11-17 | 5000.00 |    NULL |     10 |
|  7844 | TURNER | SALESMAN  | 7698 | 1981-09-08 | 1500.00 |    0.00 |     30 |
|  7876 | ADAMS  | CLERK     | 7788 | 1987-05-23 | 1100.00 |    NULL |     20 |
|  7900 | JAMES  | CLERK     | 7698 | 1981-12-03 |  950.00 |    NULL |     30 |
|  7902 | FORD   | ANALYST   | 7566 | 1981-12-03 | 3000.00 |    NULL |     20 |
|  7934 | MILLER | CLERK     | 7782 | 1982-01-23 | 1300.00 |    NULL |     10 |
+-------+--------+-----------+------+------------+---------+---------+--------+
13 rows in set (0.02 sec)

19、列出所有"CLERK"(办事员)的姓名及其部门名称,部门的人数

先找出办事员的姓名及部门编号

mysql> select ename,job, deptno from emp where job = 'clerk';
+--------+-------+--------+
| ename  | job   | deptno |
+--------+-------+--------+
| SMITH  | CLERK |     20 |
| ADAMS  | CLERK |     20 |
| JAMES  | CLERK |     30 |
| MILLER | CLERK |     10 |
+--------+-------+--------+
4 rows in set (0.00 sec)

找出每个部门所对应的编号及人数

mysql> select deptno,count(deptno) from emp group by deptno;
+--------+---------------+
| deptno | count(deptno) |
+--------+---------------+
|     10 |             3 |
|     20 |             5 |
|     30 |             6 |
+--------+---------------+
3 rows in set (0.00 sec)

再找出对应名称

mysql> select e.deptno,count(e.deptno), d.dname from emp e join dept d on e.deptno = d.deptno group by deptno;
+--------+-----------------+------------+
| deptno | count(e.deptno) | dname      |
+--------+-----------------+------------+
|     10 |               3 | ACCOUNTING |
|     20 |               5 | RESEARCH   |
|     30 |               6 | SALES      |
+--------+-----------------+------------+
3 rows in set (0.00 sec)

然后两个表连接

mysql> select a.ename, b.dname, b.cc
    -> from(
    -> select ename,job, deptno from emp where job = 'clerk') as a
    -> join(
    -> select e.deptno,count(e.deptno) as cc, d.dname from emp e join dept d on e.deptno = d.deptno group by deptno) as b
    -> on a.deptno = b.deptno;
+--------+------------+----+
| ename  | dname      | cc |
+--------+------------+----+
| SMITH  | RESEARCH   |  5 |
| ADAMS  | RESEARCH   |  5 |
| JAMES  | SALES      |  6 |
| MILLER | ACCOUNTING |  3 |
+--------+------------+----+
4 rows in set (0.03 sec)

20、列出最低薪金大于 1500 的各种工作及从事此工作的全部雇员人数.

先找出每个工作的最低薪金

mysql> select job, min(sal) from emp group by job;
+-----------+----------+
| job       | min(sal) |
+-----------+----------+
| ANALYST   |  3000.00 |
| CLERK     |   800.00 |
| MANAGER   |  2450.00 |
| PRESIDENT |  5000.00 |
| SALESMAN  |  1250.00 |
+-----------+----------+
5 rows in set (0.03 sec)

选出最低薪金大于1500的工作

mysql> select job, min(sal) from emp group by job having min(sal) > 1500;
+-----------+----------+
| job       | min(sal) |
+-----------+----------+
| ANALYST   |  3000.00 |
| MANAGER   |  2450.00 |
| PRESIDENT |  5000.00 |
+-----------+----------+
3 rows in set (0.00 sec)

从事此工作的人数

mysql> select job,count(job) from emp where job in (
    -> select job from emp group by job having min(sal) > 1500)
    -> group by job;
+-----------+------------+
| job       | count(job) |
+-----------+------------+
| ANALYST   |          2 |
| MANAGER   |          3 |
| PRESIDENT |          1 |
+-----------+------------+
3 rows in set (0.00 sec)

又写的麻烦了,在选工作的同时直接计数
mysql> select job, count(*) from emp group by job having min(sal) > 1500;

21、列出在部门"SALES"<销售部>工作的员工的姓名,假定不知道销售部的部门编号

先把两个表连起来,然后再查询就完事了

mysql> select e.*, d.dname from emp e join dept d on e.deptno = d.deptno;
+-------+--------+-----------+------+------------+---------+---------+--------+------------+
| EMPNO | ENAME  | JOB       | MGR  | HIREDATE   | SAL     | COMM    | DEPTNO | dname      |
+-------+--------+-----------+------+------------+---------+---------+--------+------------+
|  7782 | CLARK  | MANAGER   | 7839 | 1981-06-09 | 2450.00 |    NULL |     10 | ACCOUNTING |
|  7839 | KING   | PRESIDENT | NULL | 1981-11-17 | 5000.00 |    NULL |     10 | ACCOUNTING |
|  7934 | MILLER | CLERK     | 7782 | 1982-01-23 | 1300.00 |    NULL |     10 | ACCOUNTING |
|  7369 | SMITH  | CLERK     | 7902 | 1980-12-17 |  800.00 |    NULL |     20 | RESEARCH   |
|  7566 | JONES  | MANAGER   | 7839 | 1981-04-02 | 2975.00 |    NULL |     20 | RESEARCH   |
|  7788 | SCOTT  | ANALYST   | 7566 | 1987-04-19 | 3000.00 |    NULL |     20 | RESEARCH   |
|  7876 | ADAMS  | CLERK     | 7788 | 1987-05-23 | 1100.00 |    NULL |     20 | RESEARCH   |
|  7902 | FORD   | ANALYST   | 7566 | 1981-12-03 | 3000.00 |    NULL |     20 | RESEARCH   |
|  7499 | ALLEN  | SALESMAN  | 7698 | 1981-02-20 | 1600.00 |  300.00 |     30 | SALES      |
|  7521 | WARD   | SALESMAN  | 7698 | 1981-02-22 | 1250.00 |  500.00 |     30 | SALES      |
|  7654 | MARTIN | SALESMAN  | 7698 | 1981-09-28 | 1250.00 | 1400.00 |     30 | SALES      |
|  7698 | BLAKE  | MANAGER   | 7839 | 1981-05-01 | 2850.00 |    NULL |     30 | SALES      |
|  7844 | TURNER | SALESMAN  | 7698 | 1981-09-08 | 1500.00 |    0.00 |     30 | SALES      |
|  7900 | JAMES  | CLERK     | 7698 | 1981-12-03 |  950.00 |    NULL |     30 | SALES      |
+-------+--------+-----------+------+------------+---------+---------+--------+------------+
14 rows in set (0.00 sec)

mysql> select n.ename from (select e.*, d.dname from emp e join dept d on e.deptno = d.deptno) as n where dname = 'sales';
+--------+
| ename  |
+--------+
| ALLEN  |
| WARD   |
| MARTIN |
| BLAKE  |
| TURNER |
| JAMES  |
+--------+
6 rows in set (0.00 sec)

看视频后:
先根据部门名称找部门编号,然后部门编号相同找员工
select ename from emp where deptno = (select deptno from dept where dname = ‘SALES’);

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

先找薪金高于平均薪金的所有员工

mysql> select ename,deptno, mgr,sal from emp where sal > (select avg(sal) from emp);
+-------+--------+------+---------+
| ename | deptno | mgr  | sal     |
+-------+--------+------+---------+
| JONES |     20 | 7839 | 2975.00 |
| BLAKE |     30 | 7839 | 2850.00 |
| CLARK |     10 | 7839 | 2450.00 |
| SCOTT |     20 | 7566 | 3000.00 |
| KING  |     10 | NULL | 5000.00 |
| FORD  |     20 | 7566 | 3000.00 |
+-------+--------+------+---------+
6 rows in set (0.00 sec)

然后连接三个表分别查询

mysql> select e.ename, d.dname,l.ename, s.grade from (
->  select ename,deptno, mgr,sal from emp where sal > (select avg(sal) from emp)) as e
-> join dept d
-> on e.deptno = d.deptno
-> left join emp l
-> on e.mgr = l.empno
-> join salgrade s
-> on e.sal between losal and hisal;
+-------+------------+-------+-------+
| ename | dname      | ename | grade |
+-------+------------+-------+-------+
| JONES | RESEARCH   | KING  |     4 |
| BLAKE | SALES      | KING  |     4 |
| CLARK | ACCOUNTING | KING  |     4 |
| SCOTT | RESEARCH   | JONES |     4 |
| KING  | ACCOUNTING | NULL  |     5 |
| FORD  | RESEARCH   | JONES |     4 |
+-------+------------+-------+-------+
6 rows in set (0.00 sec)

看视频后:
又繁琐了

select 
	e.ename '员工',d.dname,l.ename '领导',s.grade
from
	emp e
join
	dept d
on
	e.deptno = d.deptno
left join
	emp l
on
	e.mgr = l.empno
join
	salgrade s
on
	e.sal between s.losal and s.hisal
where
	e.sal > (select avg(sal) from emp);

23、列出与"SCOTT"从事相同工作的所有员工及部门名称

找出scott从事的工作

mysql> select job from emp where ename = 'scott';
+---------+
| job     |
+---------+
| ANALYST |
+---------+
1 row in set (0.00 sec)

找到和他从事相同工作的 员工名称和部门编号

mysql> select ename,deptno from emp where job = (select job from emp where ename = 'scott') and ename != 'scott';
+-------+--------+
| ename | deptno |
+-------+--------+
| FORD  |     20 |
+-------+--------+
1 row in set (0.00 sec)

连接表查询部门名称

mysql> select e.ename, d.dname from (
-> select ename,deptno from emp where job = (select job from emp where ename = 'scott') and ename != 'scott') as e
-> join dept d
-> on e.deptno = d.deptno;
+-------+----------+
| ename | dname    |
+-------+----------+
| FORD  | RESEARCH |
+-------+----------+

看视频后:
我发现我之前做的都是一步步来,这样我就容易将每一步看成一个张表,然后再去连接另一张表进行下一步,有些地方其实不用这样麻烦
例如这里,我在找到和他从事相同工作的 员工名称和部门编号后直接连接部门表把部门名称找出来就行了

select 
	e.ename,e.job,d.dname
from
	emp e
join
	dept d
on
	e.deptno = d.deptno
where
	e.job = (select job from emp where ename = 'SCOTT')
and
	e.ename <> 'SCOTT';

24、列出薪金等于部门 30 中员工的薪金的其他员工的姓名和薪金

这句话读着咋这么别扭
先找部门30中员工的薪金及名字

mysql> select ename, sal from emp where deptno = 30;
+--------+---------+
| ename  | sal     |
+--------+---------+
| ALLEN  | 1600.00 |
| WARD   | 1250.00 |
| MARTIN | 1250.00 |
| BLAKE  | 2850.00 |
| TURNER | 1500.00 |
| JAMES  |  950.00 |
+--------+---------+
6 rows in set (0.00 sec)

找出薪资在这个表中,但名字不在这个表中的员工的姓名和薪金,
又死板了,我一直这样写
mysql> select e.ename,e.sal from emp e
-> join (
-> select ename, sal from emp where deptno = 30) as b
-> on e.sal in b.sal and e.ename not in b.ename;
但是这样不行,最下面一一行直报错
名字不在上面表中的员工,就是部门不是30的,直接筛选就完事了…

mysql> select ename,sal from emp
-> where sal in(
-> select sal from emp where deptno = 30)
-> and deptno != 30;

Empty set (0.00 sec)

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

同样先找部门30中员工的薪金的最大值

mysql> select max(sal) from emp where deptno = 30;
+----------+
| max(sal) |
+----------+
|  2850.00 |
+----------+
1 row in set (0.00 sec)

然后找大于这个薪金的员工姓名和部门编号

mysql> select ename,sal,deptno from emp
	-> where sal > (select max(sal) from emp where deptno = 30) and deptno != 30;
+-------+---------+--------+
| ename | sal     | deptno |
+-------+---------+--------+
| JONES | 2975.00 |     20 |
| SCOTT | 3000.00 |     20 |
| KING  | 5000.00 |     10 |
| FORD  | 3000.00 |     20 |
+-------+---------+--------+
4 rows in set (0.00 sec)

连接表,将部门编号变成名称

mysql> select e.ename, e.sal ,d.dname from (
    -> select ename,sal,deptno from emp
    -> where sal > (select max(sal) from emp where deptno = 30) and deptno != 30) e
    -> join dept d
    -> on e.deptno = d.deptno;
+-------+---------+------------+
| ename | sal     | dname      |
+-------+---------+------------+
| KING  | 5000.00 | ACCOUNTING |
| JONES | 2975.00 | RESEARCH   |
| SCOTT | 3000.00 | RESEARCH   |
| FORD  | 3000.00 | RESEARCH   |
+-------+---------+------------+
4 rows in set (0.00 sec)

看视频后:
我又去看成一个表了,直接连接就行了,还有不需要加部门编号不等于30
mysql> select e.ename,e.sal,d.dname from emp e join dept d on e.deptno = d.deptno
-> where sal > (select max(sal) from emp where deptno = 30);

mysql> select e.ename,e.sal,d.dname from emp  e join dept d on e.deptno = d.deptno where sal > (select max(sal) from emp where deptno = 30);
+-------+---------+------------+
| ename | sal     | dname      |
+-------+---------+------------+
| KING  | 5000.00 | ACCOUNTING |
| JONES | 2975.00 | RESEARCH   |
| SCOTT | 3000.00 | RESEARCH   |
| FORD  | 3000.00 | RESEARCH   |
+-------+---------+------------+
4 rows in set (0.00 sec)

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

不太明白这个平均服务期限是什么鬼
先写一下其他两项

mysql> select d.dname, count(e.ename),avg(e.sal) from emp e
-> join dept d
-> on e.deptno = d.deptno
-> group by e.deptno;
+------------+----------------+-------------+
| dname      | count(e.ename) | avg(e.sal)  |
+------------+----------------+-------------+
| ACCOUNTING |              3 | 2916.666667 |
| RESEARCH   |              5 | 2175.000000 |
| SALES      |              6 | 1566.666667 |
+------------+----------------+-------------+
3 rows in set (0.00 sec)

因为有一个没有员工的部门,所以这里不对,需要外连接,并对空赋值

mysql> select d.dname, count(e.ename),ifnull(avg(e.sal),0) from emp e
    -> right join dept d
    -> on e.deptno = d.deptno
    -> group by e.deptno;
+------------+----------------+----------------------+
| dname      | count(e.ename) | ifnull(avg(e.sal),0) |
+------------+----------------+----------------------+
| OPERATIONS |              0 |             0.000000 |
| ACCOUNTING |              3 |          2916.666667 |
| RESEARCH   |              5 |          2175.000000 |
| SALES      |              6 |          1566.666667 |
+------------+----------------+----------------------+
4 rows in set (0.04 sec)

服务期限,我猜是现在的时间减去入职的时间

在MySQL中计算两个日期的差值

在mysql当中怎么计算两个日期的“年差”,差了多少年?
TimeStampDiff(间隔类型, 前一个日期, 后一个日期)

timestampdiff(YEAR, hiredate, now())

间隔类型:
	SECOND   秒,
	MINUTE   分钟,
	HOUR   小时,
	DAY   天,
	WEEK   星期
	MONTH   月,
	QUARTER   季度,
	YEAR   年

mysql> select d.dname, count(e.ename) ecount,ifnull(avg(e.sal),0) avgsal,
    -> ifnull(Timestampdiff(YEAR,hiredate, now()),0) avgservetime from emp e
    -> right join dept d
    -> on e.deptno = d.deptno
    -> group by e.deptno;
+------------+--------+-------------+--------------+
| dname      | ecount | avgsal      | avgservetime |
+------------+--------+-------------+--------------+
| OPERATIONS |      0 |    0.000000 |            0 |
| ACCOUNTING |      3 | 2916.666667 |           39 |
| RESEARCH   |      5 | 2175.000000 |           40 |
| SALES      |      6 | 1566.666667 |           40 |
+------------+--------+-------------+--------------+
4 rows in set (0.02 sec)

27、列出所有员工的姓名、部门名称和工资

怎么还感觉越来越简单了

mysql> select e.ename, d.dname, e.sal
    -> from emp e
    -> join dept d
    -> on e.deptno = d.deptno;
+--------+------------+---------+
| ename  | dname      | sal     |
+--------+------------+---------+
| CLARK  | ACCOUNTING | 2450.00 |
| KING   | ACCOUNTING | 5000.00 |
| MILLER | ACCOUNTING | 1300.00 |
| SMITH  | RESEARCH   |  800.00 |
| JONES  | RESEARCH   | 2975.00 |
| SCOTT  | RESEARCH   | 3000.00 |
| ADAMS  | RESEARCH   | 1100.00 |
| FORD   | RESEARCH   | 3000.00 |
| ALLEN  | SALES      | 1600.00 |
| WARD   | SALES      | 1250.00 |
| MARTIN | SALES      | 1250.00 |
| BLAKE  | SALES      | 2850.00 |
| TURNER | SALES      | 1500.00 |
| JAMES  | SALES      |  950.00 |
+--------+------------+---------+
14 rows in set (0.00 sec)

28、列出所有部门的详细信息和人数

mysql> select d.*,ifnull(e.ecount,0) from (
    -> select deptno,count(ename) ecount from emp group by deptno) as e
    -> right join dept d
    -> on e.deptno = d.deptno;
+--------+------------+----------+--------------------+
| DEPTNO | DNAME      | LOC      | ifnull(e.ecount,0) |
+--------+------------+----------+--------------------+
|     10 | ACCOUNTING | NEW YORK |                  3 |
|     20 | RESEARCH   | DALLAS   |                  5 |
|     30 | SALES      | CHICAGO  |                  6 |
|     40 | OPERATIONS | BOSTON   |                  0 |
+--------+------------+----------+--------------------+
4 rows in set (0.00 sec)

又复杂了

select 
	d.deptno,d.dname,d.loc,count(e.ename)
from
	emp e
right join
	dept d
on
	e.deptno = d.deptno
group by
	d.deptno,d.dname,d.loc;

29、列出各种工作的最低工资及从事此工作的雇员姓名

首先,各种工作的最低工资和名称

mysql> select job,min(sal) from emp group by job;
+-----------+----------+
| job       | min(sal) |
+-----------+----------+
| ANALYST   |  3000.00 |
| CLERK     |   800.00 |
| MANAGER   |  2450.00 |
| PRESIDENT |  5000.00 |
| SALESMAN  |  1250.00 |
+-----------+----------+
5 rows in set (0.00 sec)

根据所给出的答案判断,题目应该是从事此工作的并且工资是最低的雇员姓名

mysql> select e.ename,j.job,j.minsal from emp e
    -> join (
    -> select job,min(sal) minsal from emp group by job) as j
    -> on e.job = j.job and e.sal = j.minsal;
+--------+-----------+---------+
| ename  | job       | minsal  |
+--------+-----------+---------+
| SMITH  | CLERK     |  800.00 |
| WARD   | SALESMAN  | 1250.00 |
| MARTIN | SALESMAN  | 1250.00 |
| CLARK  | MANAGER   | 2450.00 |
| SCOTT  | ANALYST   | 3000.00 |
| KING   | PRESIDENT | 5000.00 |
| FORD   | ANALYST   | 3000.00 |
+--------+-----------+---------+
7 rows in set (0.00 sec)

30、列出各个部门的 MANAGER(领导)的最低薪金

这里特别注意,先判断是领导,再分组,可以得到答案;但是先分组,再判断会显示没有job这一列
因此可以得到结论,分组以后,表中所剩的字段就只剩下分组所用字段和分组函数了,然后才会去执行having

mysql> select deptno, min(sal) from emp group by deptno having job = 'manager';
ERROR 1054 (42S22): Unknown column 'job' in 'having clause'

mysql> select deptno, min(sal) from emp where job= 'manager' group by deptno;
+--------+----------+
| deptno | min(sal) |
+--------+----------+
|     10 |  2450.00 |
|     20 |  2975.00 |
|     30 |  2850.00 |
+--------+----------+
3 rows in set (0.00 sec)

31、列出所有员工的年工资,按年薪从低到高排序

mysql> select ename,(sal * 12 + ifnull(comm, 0) * 12) yearsal from emp order by yearsal asc;
+--------+----------+
| ename  | yearsal  |
+--------+----------+
| SMITH  |  9600.00 |
| JAMES  | 11400.00 |
| ADAMS  | 13200.00 |
| MILLER | 15600.00 |
| TURNER | 18000.00 |
| WARD   | 21000.00 |
| ALLEN  | 22800.00 |
| CLARK  | 29400.00 |
| MARTIN | 31800.00 |
| BLAKE  | 34200.00 |
| JONES  | 35700.00 |
| FORD   | 36000.00 |
| SCOTT  | 36000.00 |
| KING   | 60000.00 |
+--------+----------+
14 rows in set (0.00 sec)

32、求出员工领导的薪水超过 3000 的员工名称与领导名称

mysql> select e.ename, l.ename,l.sal
    -> from emp e
    -> join emp l
    -> on e.mgr = l.empno
    -> where l.sal > 3000;
+-------+-------+---------+
| ename | ename | sal     |
+-------+-------+---------+
| JONES | KING  | 5000.00 |
| BLAKE | KING  | 5000.00 |
| CLARK | KING  | 5000.00 |
+-------+-------+---------+
3 rows in set (0.00 sec)

33、求出部门名称中,带’S’字符的部门员工的工资合计、部门人数

模糊查询用一下,求出带S字符的部门名称

mysql> select deptno,dname from dept where dname like '%s%';
+--------+------------+
| deptno | dname      |
+--------+------------+
|     20 | RESEARCH   |
|     30 | SALES      |
|     40 | OPERATIONS |
+--------+------------+
3 rows in set (0.00 sec)

然后连接表查工资合计和部门人数

mysql> select d.deptno, d.dname, sum(e.sal), count(e.ename)
    -> from emp e
    -> right join (
    -> select deptno,dname from dept where dname like '%s%') as d
    -> on e.deptno = d.deptno
    -> group by deptno;
+--------+------------+------------+----------------+
| deptno | dname      | sum(e.sal) | count(e.ename) |
+--------+------------+------------+----------------+
|     20 | RESEARCH   |   10875.00 |              5 |
|     30 | SALES      |    9400.00 |              6 |
|     40 | OPERATIONS |       NULL |              0 |
+--------+------------+------------+----------------+
3 rows in set (0.00 sec)

看视频后:
也不用嵌套,

select 
	d.deptno,d.dname,d.loc,count(e.ename),ifnull(sum(e.sal),0) as sumsal
from
	emp e
right join
	dept d
on
	e.deptno = d.deptno
where
	d.dname like '%S%'
group by
	d.deptno,d.dname,d.loc;

34、给任职日期超过 30 年的员工加薪 10%

全超过30年了哈哈

mysql> select ename, sal * 1.1 from emp where timestampdiff(year, hiredate, now()) > 30;
+--------+-----------+
| ename  | sal * 1.1 |
+--------+-----------+
| SMITH  |    880.00 |
| ALLEN  |   1760.00 |
| WARD   |   1375.00 |
| JONES  |   3272.50 |
| MARTIN |   1375.00 |
| BLAKE  |   3135.00 |
| CLARK  |   2695.00 |
| SCOTT  |   3300.00 |
| KING   |   5500.00 |
| TURNER |   1650.00 |
| ADAMS  |   1210.00 |
| JAMES  |   1045.00 |
| FORD   |   3300.00 |
| MILLER |   1430.00 |
+--------+-----------+
14 rows in set (0.03 sec)

总结

花了两三天把这些题做完了,刚开始做的时候有点懵,后面知道套路以后就越做越快了,当然也和后面的题变简单有一定关系,当然也发现了一些小问题,感觉数据库还是比较简单的,继续加油了

看了视频,发现自己很多题都是一步一步来的,这样虽然很条理,不容易出错,但是有很多过程中我都是把上一步结果变成了一张表,然后嵌套处理,其实都没必要,所以这里注意一下

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