已知2张基本表：部门表：dept （部门号，部门名称）;员工表 emp（员工号，员工姓名，年龄，入职时间，收入，部门号）
1：dept表中有4条记录：
部门号(dept1) 部门名称(dept_name )
101 财务
102 销售
103 IT技术
104 行政
2：emp表中有6条记录：
员工号 员工姓名 年龄 入职时间 收入 部门号对应字段名称为： (sid name age worktime_start incoming dept2)
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
1.列出每个部门的平均收入及部门名称;
方法一;
1) SELECT AVG(incoming) dept_name FROM emp RIGHT JOIN dept ON emp.dept2 = dept.dept1 GROUP BY dept_name;
方法二：
2)select s.dept_name, m.n from dept s left join (select avg(incoming)n ,dept2 from emp group by dept2 ) m on s.dept1=m.dept2;
2.财务部门的收入总和；
SELECT dept_name ,SUM(incoming) FROM emp INNER JOIN dept ON dept.dept1 = emp.dept2 WHERE dept_name = "财务";
2)select sum(incoming) from emp where dept2=(select dept1 from dept where dept_name="财务");
3.It技术部入职员工的员工号
SELECT dept_name ,sid FROM emp INNER JOIN dept ON dept.dept1 = emp.dept2 WHERE dept_name = "IT技术";
2)select emp.sid from dept inner join emp on dept.dept1=emp.dept2 where dept_name="技术 ";
4.财务部门收入超过2000元的员工姓名
SELECT name ,incoming FROM emp INNER JOIN dept ON dept.dept1 = emp.dept2 WHERE dept_name = "财务" AND incoming >2000;
select name from dept INNER JOIN emp on dept.dept1=emp.dept2 where dept_name=”财务 ” and incoming>2000；
5.找出销售部收入最低的员工的入职时间;
SELECT woektime_start
FROM
(SELECT * FROM emp INNER JOIN dept ON emp.dept2 = dept.dept1 ORDER BY incoming)a
WHERE dept_name = "销售"
LIMIT 1;
select s.woektime_start from (select woektime_start,min(incoming) from (select * from emp INNER JOIN dept on dept.dept1=emp.dept2)q where q.dept_name="销售")s ;
方法三
select woektime_start from emp,dept where dept.dept1=emp.dept2 and
incoming in (select min(incoming) from emp where dept2=(select dept1 from dept where dept_name="销售")) and dept_name ="销售" ;
6.找出年龄小于平均年龄的员工的姓名，ID和部门名称
SELECT name,sid,dept_name FROM emp INNER JOIN dept ON dept.dept1 = emp.dept2 where age< (SELECT AVG(age) FROM emp);
2)select dept_name,sid,name from emp INNER JOIN dept on dept1=emp.dept2 and age < (select avg(age) from emp join dept on dept1=emp.dept2);
7.列出每个部门收入总和高于9000的部门名称
SELECT
dept_name
FROM
( SELECT sum( incoming ) a, dept_name FROM emp INNER JOIN dept ON dept.dept1 = emp.dept2 GROUP BY dept_name ) q
WHERE
a > 9000;
select dept.dept_name from dept inner join ( select sum(incoming)s , dept2 from emp group by dept2 having s>9000) c on dept.dept1=c.dept2 ;
方法三：
select dept_name from emp left join dept on emp.dept2=dept.dept1 group by dept_name having sum(incoming)>9000;
8.查出财务部门工资少于3800元的员工姓名
SELECT NAME
FROM
( SELECT name,dept_name,incoming FROM emp INNER JOIN dept ON dept.dept1 = emp.dept2 ) a
WHERE
dept_name = "财务"
AND incoming < 3800;
9.求财务部门最低工资的员工姓名；
SELECT NAME
FROM
( SELECT * FROM emp INNER JOIN dept ON dept.dept1 = emp.dept2 WHERE dept_name = "财务" ORDER BY incoming) a
LIMIT 1;
select name from dept inner join emp on dept.dept1=emp.dept2 where dept2=(select dept1 from dept where dept_name=”财务”and incoming=(select min(incoming) from emp ,dept where emp.dept2=dept.dept1 and dept_name="财务");
10.找出销售部门中年纪最大的员工的姓名
SELECT NAME FROM
( SELECT * FROM emp INNER JOIN dept ON dept.dept1 = emp.dept2 WHERE dept_name = "销售" ORDER BY age DESC) a
LIMIT 1;
select name from emp join dept on dept1=emp.dept2 where dept_name="销售" and age=( select max(age) from emp join dept on dept1=emp.dept2 where dept_name="销售")
11.求收入最低的员工姓名及所属部门名称：
SELECT name ,dept_name FROM(SELECT * FROM emp INNER JOIN dept ON dept.dept1 = emp.dept2)a WHERE incoming =( SELECT MIN(incoming) FROM emp);
//方法二
SELECT NAME,
dept_name
FROM
( SELECT * FROM emp INNER JOIN dept ON dept.dept1 = emp.dept2 ORDER BY incoming ) a
LIMIT 1;
12.求李四的收入及部门名称
条件： name =“李四”
结果：incoming ，dept_name
方法一：
SELECT incoming,dept_name FROM emp INNER JOIN dept ON dept.dept1 = emp.dept2 WHERE NAME = "李四";
方法二：
select dept_name,incoming from dept left join emp on dept1=dept2 where name="李四" ；
方法三：
select s.incoming ,dept_name from dept,(select incoming, dept2 from emp where name='李四')s where dept.dept1=s.dept2 ;
方法四：
select s.dept_name , s.incoming from (
select * from dept left join emp on dept1=dept2 )s where s.name="李四" ;
13.求员工收入小于4000元的员工部门编号及其部门名称
条件：incoming<4000
结果：dept1, dept_name
方法一：SELECT dept1,dept_name FROM ( SELECT * FROM emp INNER JOIN dept ON dept.dept1 = emp.dept2 ) a WHERE incoming < 4000;
方法二：select dept_name ,dept1 from emp ,dept where dept1=emp.dept2 and incoming < 4000;
14.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；
第一种方法：SELECT name , dept_name,incoming FROM
( SELECT * FROM emp RIGHT JOIN dept ON dept.dept1 = emp.dept2 ORDER BY incoming DESC ) a
GROUP BY dept_name ORDER BY incoming DESC;
第二种方法：
diyuSELECT name,incoming,dept_name from
(SELECT name,dept_name,incoming,dept2 from emp right JOIN dept on dept.dept1=emp.dept2 ORDER BY incoming DESC) b
GROUP BY
dept2
ORDER BY
incoming desc;
15.求出财务部门收益最高的俩位员工的姓名，工号，收益
条件： dept_name="财务" ， max（incoming）或排序方法取值
结果：name，sid ，incoming
方法一：先临时表里面合表在排序（降序），在从临时表中取值两行
理解：可以认为在一个单表中查询数据（单表结构（单表中查询的是多表，所以要先合表））
SELECT name , sid,incoming FROM ( SELECT * FROM emp INNER JOIN dept ON dept.dept1 = emp.dept2 WHERE dept_name = "财务" ORDER BY incoming DESC ) a LIMIT 2;
方法二:先合表，在排序（降序），在取值两行
select name,sid,incoming from dept inner join emp on dept1=dept2 where dept_name='财务' order by incoming desc limit 2;
方法三：先子查询把财务部门的编号求出；在用emp表dept2关联子查询中dept1，求出我们结果
子查询
select name,sid,incoming from emp where dept2=(select dept1 from dept where dept_name='财务') order by incoming desc limit 2;
16.查询财务部低于平均收入的员工号与员工姓名：(平均工资指全体平均工资)
条件：dept_name="财务部" ， avg（*） 或avg（incomg）或avg（1） dept
结果：sid ，name emp
方法一：
SELECT a.name,a.sid FROM (SELECT name ,sid,incoming FROM emp INNER JOIN dept ON emp.dept2 = dept.dept1 WHERE dept_name = "财务")a WHERE incoming <(SELECT AVG(incoming) FROM emp);
方法二：
select name,sid from dept left join emp on dept1=dept2 where incoming<(select avg(incoming)from dept left join emp on dept1=dept2) and dept_name="财务";
方法三：
select sid,name from dept,emp where dept1=dept2 and incoming<(select avg(incoming) from emp) and dept_name='财务'
方法四：
select sid,name from emp where incoming<(select avg(incoming)from emp) and dept2=(select dept1 from dept where dept_name='财务');
17.列出部门员工数大于1个的部门名称；
方法一：
SELECT dept_name ,n FROM (SELECT COUNT(dept_name) as n,dept_name FROM emp INNER JOIN dept ON emp.dept2 = dept.dept1 GROUP BY dept_name )a WHERE n >1;
方法二： SELECT dept_name from dept LEFT JOIN emp on dept.dept1=emp.dept2 GROUP BY dept_name HAVING COUNT(dept2)>1;
方法三：select dept_name from dept where dept1 in (select dept2 from emp group by dept2 having count(dept2)>1);
18.列出部门员工收入不超过7500，且大于3000的员工年纪及部门编号；
条件： dept_nem incoming<7500 incoming>3000
结果：age ,dept2
方法一：
SELECT age,dept1 FROM (SELECT * FROM emp INNER JOIN dept ON emp.dept2 = dept.dept1 )a
WHERE incoming <=7500 and incoming >3000;
方法二：
SELECT age,dept2 from emp WHERE incoming>'3000' and incoming<='7500'
方法三：
select age,dept2 from emp inner join dept on dept.dept1=emp.dept2 where incoming between 3000 and 7500;
19.求入职于20世纪70年代的员工所属部门名称；
条件： 20世纪70年代 197%
结果：dept_name
方法一：
SELECT dept_name FROM (SELECT * FROM emp INNER JOIN dept ON emp.dept2 = dept.dept1 )a
WHERE woektime_start LIKE "197%";
方法二：
SELECT dept_name FROM emp right JOIN dept on dept.dept1=emp.dept2 where woektime_start BETWEEN 1970 and 1979;
方法三:
SELECT dept_name FROM emp right JOIN dept on dept.dept1=emp.dept2 where woektime_start > 1969 and woektime_star <1980;
方法四：
20.查找张三所在的部门名称；
条件： name =“张三”
结果：dept_name
方法一：SELECT dept_name FRO (SELECT * FROM emp INNER JOIN dept ON emp.dept2 = dept.dept1 )a WHERE name = "张三";
方法二：select dept_name from dept left join emp on dept.dept1=emp.dept2 where name=”张三
方法三：select dept_name from dept where dept1=(select dept2 from emp where name= "张三")
21.列出每一个部门中年纪最大的员工姓名，部门名称；
条件：每个部门 group max（age）
结果：name ，dept_name
方法一：
SELECT dept_name,name
FROM
(SELECT * FROM emp INNER JOIN dept ON emp.dept2 = dept.dept1 ORDER BY age desc )a
GROUP BY dept_name;
SELECT
MAX(age),dept_name,`name` from emp RIGHT JOIN dept on dept.dept1=emp.dept2
GROUP BY
dept_name;
方法二：
select dept_name ,name from (select dept_name,name ,age from dept left join emp on dept1=dept2 where age in (select max(age) from dept left join emp on dept1=dept2 group by dept_name) order by age DESC)s group by dept_name ;
方法三：错误方法
select dept_name ,name from (select dept_name,name ,age from dept left join emp on dept1=dept2 where age in (select max(age) from dept left join emp on dept1=dept2 group by dept_name) )s group by dept_name having max(age) ;
方法四：
select t.dept_name ,t.name from ( select dept_name ,name ,max(age) from (select dept_name,name ,age from dept left join emp on dept1=dept2 where age in (select max(age) c from dept left join emp on dept1=dept2 group by dept_name) )s group by dept_name )t ;
在解答：
22.列出每一个部门的员工总收入及部门名称；
条件：group by sum（incoming）
结果：sum（incoming） ，dept_name
方法一：
SELECT SUM(incoming) as a,dept_name FROM emp INNER JOIN dept ON emp.dept2 = dept.dept1 GROUP BY dept_name;
方法二：select s.a,dept.dept_name from dept left join (select sum(incoming) a,dept2 from emp group by dept2
)s on dept.dept1=s.dept2 ;
23.列出部门员工收入大于7000的员工号，部门名称；
条件： incoming> 7000
结果：sid ,dept_name
方法一：
SELECT a.sid,a.dept_name FROM (SELECT * FROM emp INNER JOIN dept ON emp.dept2 = dept.dept1 WHERE incoming >7000 )a;
方法二：
select sid,dept_name FROM(SELECT * fROM emp INNER JOIN dept ON emp.dept2 = dept.dept1)a where incoming>7000
方法三：
SELECT sid,dept_name FROM emp INNER JOIN dept ON emp.dept2 = dept.dept1 WHERE incoming >7000 ；
24.找出哪个部门还没有员工入职；
条件：dept_name name
方法一：
SELECT dept_name FROM (SELECT * FROM emp RIGHT JOIN dept ON emp.dept2 = dept.dept1 )a
WHERE sid IS NULL;
方法二：（所有的另一个表为空）
select dept_name from dept left join emp on dept1=dept2 where sid is null and name is null and age is null and woektime_start is null and incoming is null;
方法三：
SELECT * FROM dept left JOIN emp ON emp.dept2 = dept.dept1 where name is null；
方法四：
select dept_name from dept where dept1 not in (select dept2 from emp);
方法五：
select dept_name from dept where dept1 !=all (select dept2 from emp);
25.先按部门号大小排序，再依据入职时间由早到晚排序员工信息表 ；
方法一：
SELECT * FROM emp INNER JOIN dept ON emp.dept2 = dept.dept1 ORDER BY dept1 DESC,
woektime_start ASC;
方法二：
select * from emp order by dept2 desc ,woektime_start asc ;
方法三：错误方法
select*from (select*from dept left join emp on dept1=dept2 order by dept1 desc)a order by woektime_start asc;
26.求出财务部门工资最高员工的姓名和员工号
第一方法：（如果重复最高，只能取一个）
SELECT name,sid FROM (SELECT * FROM emp INNER JOIN dept ON emp.dept2 = dept.dept1 WHERE dept_name = "财务" ORDER BY incoming DESC)a LIMIT 0,1
第二方法：（更精确）
select name ,sid from emp join dept on dept1=emp.dept2 where dept_name="财务" and incoming=( select max(incoming) from emp join dept on dept1=emp.dept2 where dept_name="财务")
27.求出工资在7500到8500之间，年龄最大的员工的姓名和部门名
第一方法：
SELECT name,dept_name FROM(SELECT * FROM emp INNER JOIN dept ON emp.dept2 = dept.dept1 ORDER BY age DESC)a WHERE incoming>=7500 AND incoming <=8500 LIMIT 1;
第二种方法：
select dept_name ,name from emp,dept WHERE dept.dept1=emp.dept2 AND age in (select MAX(age) from emp where incoming between 7500 and 8500 ) and incoming between 7500 and 8500 ;
错误写法一：
select dept_name,name from dept left join emp on dept1=dept2 where incoming between 7500 and 8500 and age in(select max(age) from dept left join emp on dept1=dept2 group by dept_name);
归纳：
固定套路：
SELECT
需要展示的字段
FROM
（SELECT * FROM ）a包含全部字段的联合表，并进行分组以及排序将联合表
WHERE
条件字段;
#建表
CREATE table dept1(dept1 VARCHAR(6),dept_name VARCHAR(20));
INSERT into dept VALUES ('101','财务');
-- INSERT into dept VALUES ('102','销售');
-- INSERT into dept VALUES ('103','IT技术');
-- INSERT into dept VALUES ('104','行政');
-- INSERT into dept VALUES ('104','销售');
CREATE table emp (sid VARCHAR(20),name VARCHAR(20),age TINYINT(2),woektime_start VARCHAR(10),incoming SMALLINT(10),dept2 VARCHAR(6));
insert into emp VALUES ('1789','张三',35,'1980/1/1',4000,'101');
insert into emp VALUES ('1674','李四',32,'1983/4/1',3500,'101');
insert into emp VALUES ('1776','王五',24,'1990/7/1',2000,'101');
insert into emp VALUES ('1568','赵六',57,'1970/10/11',7500,'102');
insert into emp VALUES ('1564','荣七',64,'1963/10/11',8500,'102');
insert into emp VALUES ('1879','牛八',55,'1971/10/20',7300,'103');