MySQL——第五次作业

新增员工表emp和部门表dept
 create table dept (dept1 int ,dept_name varchar(11)) charset=utf8;
 create table emp (sid int ,name varchar(11),age int,worktime_start date,incoming int,dept2 int) charset=utf8;

     insert into dept values
    (101,'财务'),
    (102,'销售'),
    (103,'IT技术'),
    (104,'行政');

     insert into emp values
    (1789,'张三',35,'1980/1/1',4000,101),
    (1674,'李四',32,'1983/4/1',3500,101),
    (1776,'王五',24,'1990/7/1',2000,101),
    (1568,'赵六',57,'1970/10/11',7500,102),
    (1564,'荣七',64,'1963/10/11',8500,102),
    (1879,'牛八',55,'1971/10/20',7300,103);
    (1668, '钱九', 64, '1963/5/4', 8000, 102);
    (1724, '武十', 22, '2023/5/8', 1500, 103);
    (1770, '孙二', 65, '1986/8/12', 9500, 101);
    (18400, '苟一', 65, '1986/8/12', 1500, 101);

mysql> select * from emp;
+-------+------+------+----------------+----------+-------+
| sid   | name | age  | worktime_start | incoming | dept2 |
+-------+------+------+----------------+----------+-------+
|  1789 | 张三 |   35 | 1980-01-01     |     4000 |   101 |
|  1674 | 李四 |   32 | 1983-04-01     |     3500 |   101 |
|  1776 | 王五 |   24 | 1990-07-01     |     2000 |   101 |
|  1568 | 赵六 |   57 | 1970-10-11     |     7500 |   102 |
|  1564 | 荣七 |   64 | 1963-10-11     |     8500 |   102 |
|  1879 | 牛八 |   55 | 1971-10-20     |     7300 |   103 |
|  1668 | 钱九 |   64 | 1963-05-04     |     8000 |   102 |
|  1724 | 武十 |   22 | 2023-05-08     |     1500 |   103 |
|  1770 | 孙二 |   65 | 1986-08-12     |     9500 |   101 |
| 18400 | 苟一 |   65 | 1986-08-12     |     1500 |   101 |
+-------+------+------+----------------+----------+-------+
10 rows in set (0.00 sec)

mysql> select * from dept;
+-------+-----------+
| dept1 | dept_name |
+-------+-----------+
|   101 | 财务      |
|   102 | 销售      |
|   103 | IT技术    |
|   104 | 行政      |
+-------+-----------+
4 rows in set (0.00 sec)

1.找出销售部门中年纪最大的员工的姓名

mysql> select age maxage,group_concat(name) name from emp
    -> where dept2 = (select dept1 from dept where dept_name='销售')
    -> group by age
    -> order by age desc limit 1;
+--------+-----------+
| maxage | name      |
+--------+-----------+
|     64 | 荣七,钱九 |
+--------+-----------+
1 row in set (0.00 sec)

 2.求财务部门最低工资的员工姓名

mysql> select incoming min_incoming,group_concat(name) name from emp
    -> where dept2 = (select dept1 from dept where dept_name='财务')
    -> group by incoming
    -> order by incoming limit 1;
+--------------+------+
| min_incoming | name |
+--------------+------+
|         1500 | 苟一 |
+--------------+------+
1 row in set (0.00 sec)

 3.列出每个部门收入总和高于9000的部门名称

mysql> select dept_name from dept d,emp e
    -> where d.dept1=e.dept2
    -> group by dept_name
    -> having sum(incoming)>9000;
+-----------+
| dept_name |
+-----------+
| 财务      |
| 销售      |
+-----------+
2 rows in set (0.00 sec)

4.求工资在7500到8500元之间，年龄最大的人的姓名及部门

mysql> select e.name,d.dept_name from (select * from emp where incoming between 7500 and 8500) e
    -> join dept d on
    -> e.dept2=d.dept1
    -> order by age desc limit 1;
+------+-----------+
| name | dept_name |
+------+-----------+
| 钱九 | 销售      |
+------+-----------+
1 row in set (0.00 sec)

 5.找出销售部门收入最低的员工入职时间

mysql> select incoming,group_concat(worktime_start) from emp e,dept d
    -> where e.dept2 = d.dept1
    -> group by incoming
    -> order by incoming limit 1;
+----------+------------------------------+
| incoming | group_concat(worktime_start) |
+----------+------------------------------+
|     1500 | 1986-08-12,2023-05-08        |
+----------+------------------------------+
1 row in set (0.00 sec)

6.财务部门收入超过2000元的员工姓名

mysql> select name from (select * from dept where dept_name='财务') d
    -> join emp e on d.dept1 = e.dept2
    -> where incoming>2000;
+------+
| name |
+------+
| 张三 |
| 李四 |
| 孙二 |
+------+
3 rows in set (0.00 sec)

7.列出每个部门的平均收入及部门名称

mysql> select dept_name,avg(incoming) from emp e,dept d
    -> where e.dept2 = d.dept1
    -> group by dept_name;
+-----------+---------------+
| dept_name | avg(incoming) |
+-----------+---------------+
| 财务      |     4100.0000 |
| 销售      |     8000.0000 |
| IT技术    |     4400.0000 |
+-----------+---------------+
3 rows in set (0.00 sec)

  8.IT技术部入职员工的员工号

mysql> select name,sid,dept_name from dept d,emp e
    -> where d.dept1=e.dept2
    -> and dept_name='IT技术';
+------+------+-----------+
| name | sid  | dept_name |
+------+------+-----------+
| 牛八 | 1879 | IT技术    |
| 武十 | 1724 | IT技术    |
+------+------+-----------+
2 rows in set (0.00 sec)

9.财务部门的收入总和；

mysql> select dept_name,sum(incoming) from emp e,dept d
    -> where d.dept1=e.dept2
    -> group by dept_name
    -> having dept_name='财务';
+-----------+---------------+
| dept_name | sum(incoming) |
+-----------+---------------+
| 财务      |         20500 |
+-----------+---------------+
1 row in set (0.00 sec)

  10.找出哪个部门还没有员工入职；

mysql> select dept_name from emp e right join dept d on
    -> e.dept2 = d.dept1
    -> where e.sid is null;
+-----------+
| dept_name |
+-----------+
| 行政      |
+-----------+
1 row in set (0.00 sec)

  11.列出部门员工收入大于7000的部门编号，部门名称；

mysql> select distinct dept1,dept_name from(select * from emp where incoming>7000) e join dept d
    -> on e.dept2 = d.dept1;
+-------+-----------+
| dept1 | dept_name |
+-------+-----------+
|   101 | 财务      |
|   102 | 销售      |
|   103 | IT技术    |
+-------+-----------+
3 rows in set (0.00 sec)

12.列出每一个部门的员工总收入及部门名称；

mysql> select dept_name,sum(incoming) from emp e,dept d
    -> where e.dept2 = d.dept1
    -> group by dept_name;
+-----------+---------------+
| dept_name | sum(incoming) |
+-----------+---------------+
| 财务      |         20500 |
| 销售      |         24000 |
| IT技术    |          8800 |
+-----------+---------------+
3 rows in set (0.00 sec)

13.列出每一个部门中年纪最大的员工姓名，部门名称；

mysql> select e.`name`, d.dept_name from emp e join dept d on e.dept2=d.dept1 where e.age=(select max(e1.age) from emp e1 where e1.dept2=e.dept2);
+------+-----------+
| name | dept_name |
+------+-----------+
| 荣七 | 销售      |
| 牛八 | IT技术    |
| 钱九 | 销售      |
| 孙二 | 财务      |
| 苟一 | 财务      |
+------+-----------+
5 rows in set (0.00 sec)

  14.求李四的收入及部门名称

mysql> select incoming,dept_name from(select * from emp where name='李四') e join dept d
    -> on e.dept2=d.dept1;
+----------+-----------+
| incoming | dept_name |
+----------+-----------+
|     3500 | 财务      |
+----------+-----------+
1 row in set (0.00 sec)

15.列出每个部门中收入最高的员工姓名，部门名称，收入，并按照收入降序

mysql> select e.'name',d.dept_name,e.incoming from emp e join dept d on e.dept2=d.dept1
    -> where e.incoming = (select max(e1.incoming) from emp e1 where e1.dept2=e.dept2)
    -> order by e.incoming desc;