sql语句实例练习

1.最晚入职员工查询

select * from employees

where hire_date =

(select max(hire_date) from employees)

 

2.倒数第三名入职信息

select * from employees order by hire_date desc limit 2,1;

 

3.查找各个部门当前(to_date='9999-01-01')领导当前薪水详情以及其对应部门编号dept_no

select salaries.emp_no,salaries.salary,salaries.from_date,salaries.to_date,dept_manager.dept_no

from salaries inner join dept_manager

on dept_manager.emp_no = salaries.emp_no

and dept_manager.to_date = '9999-01-01'

and salaries.to_date = '9999-01-01';

4.查找所有已经分配部门的员工的last_name和first_name

select employees.last_name, first_name, dept_emp.dept_no from employees inner join dept_emp on employees.emp_no=dept_emp.emp_no

5.查找所有员工的last_name和first_name以及对应部门编号dept_no，也包括展示没有分配具体部门的员工

select a.last_name,a.first_name,b.dept_no

from employees a

left join dept_emp b

on a.emp_no=b.emp_no

6.查找所有员工入职时候的薪水情况，给出emp_no以及salary， 并按照emp_no进行逆序

select employees.emp_no,salaries.salary

from employees,salaries

where employees.emp_no=salaries.emp_no

and employees.hire_date=salaries.from_date

order by employees.emp_no desc;

7.查找薪水涨幅超过15次的员工号emp_no以及其对应的涨幅次数

select a.* from(

select emp_no,count(*) as t

from salaries

group by emp_no) as a

where t>15

8.找出所有员工当前(to_date='9999-01-01')具体的薪水salary情况，对于相同的薪水只显示一次,并按照逆序显示

select salary from salaries where to_date='9999-01-01' group by salary order by salary DESC

9.获取所有部门当前manager的当前薪水情况，给出dept_no, emp_no以及salary，当前表示to_date='9999-01-01

SELECT dept_manager.dept_no,dept_manager.emp_no,salaries.salary

from salaries,dept_manager

where dept_manager.to_date = '9999-01-01'

and salaries.to_date = '9999-01-01'

and dept_manager.emp_no = salaries.emp_no;

10.获取所有非manager的员工emp_no

select e.emp_no

from employees e left join dept_manager d

on e.emp_no = d.emp_no

where d.emp_no is null;

11.获取所有部门中当前员工薪水最高的相关信息，给出dept_no, emp_no以及其对应的salary

SELECT a.dept_no,a.emp_no,max(b.salary) salary FROM

dept_emp a,salaries b

WHERE a.to_date='9999-01-01'

AND b.to_date='9999-01-01'

AND a.emp_no=b.emp_no

GROUP BY a.dept_no

;

12.从titles表获取按照title进行分组，每组个数大于等于2，给出title以及对应的数目t

SELECT title,COUNT(title) AS t

FROM titles

GROUP BY title

HAVING t > 1;

13.从titles表获取按照title进行分组，每组个数大于等于2，给出title以及对应的数目t。（注意对于重复的emp_no进行忽略）

select title, count(distinct emp_no) as t

from titles

group by title

having t >= 2

;

14.查找employees表所有emp_no为奇数，且last_name不为Mary的员工信息，并按照hire_date逆序排列

select * from employees

where emp_no % 2 = 1

and last_name != 'Mary'

order by hire_date desc

15.统计出当前各个title类型对应的员工当前薪水对应的平均工资。结果给出title以及平均工资avg

select title,avg(salary) as avg

from titles a

inner join salaries b

on a.emp_no=b.emp_no and a.to_date = '9999-01-01' AND b.to_date = '9999-01-01'

group by a.title

16.获取当前（to_date='9999-01-01'）薪水第二多的员工的emp_no以及其对应的薪水salary

select emp_no,salary from salaries s where s.to_date = '9999-01-01' order by s.salary desc limit 1,1

17.查找当前薪水(to_date='9999-01-01')排名第二多的员工编号emp_no、薪水salary、last_name以及first_name，不准使用order by

select e.emp_no, max(s.salary) AS salary, e.last_name, e.first_name

from employees AS e inner join salaries AS s on e.emp_no=s.emp_no

where to_date='9999-01-01'and salary not in

(select max (salary) from salaries where to_date='9999-01-01')

18.查找所有员工的last_name和first_name以及对应的dept_name，也包括暂时没有分配部门的员工SELECT

e.last_name, e.first_name, dp.dept_name FROM employees AS e

LEFT JOIN dept_emp AS d ON e.emp_no = d.emp_no

LEFT JOIN departments AS dp ON d.dept_no = dp.dept_no

19.查找员工编号emp_no为10001其自入职以来的薪水salary涨幅值growth

SELECT (

(SELECT salary FROM salaries WHERE emp_no = 10001 ORDER BY to_date DESC LIMIT 1)

- (SELECT salary FROM salaries WHERE emp_no = 10001 ORDER BY to_date LIMIT 1)

) AS growth;

20.查找所有员工自入职以来的薪水涨幅情况，给出员工编号emp_no以及其对应的薪水涨幅growth，并按照growth进行升序

SELECT sCurrent.emp_no, (sCurrent.salary-sStart.salary) AS growth

FROM (SELECT s.emp_no, s.salary FROM employees e, salaries s WHERE e.emp_no = s.emp_no AND s.to_date = '9999-01-01') AS sCurrent,

(SELECT s.emp_no, s.salary FROM employees e, salaries s WHERE e.emp_no = s.emp_no AND s.from_date = e.hire_date) AS sStart

WHERE sCurrent.emp_no = sStart.emp_no

ORDER BY growth

21.统计各个部门对应员工涨幅的次数总和，给出部门编码dept_no、部门名称dept_name以及次数sum

select d.dept_no ,dept.dept_name,count(salary) from salaries s ,dept_emp d,departments dept

where s.emp_no = d.emp_no and d.dept_no=dept.dept_no group by dept.dept_no

22.对所有员工的当前(to_date='9999-01-01')薪水按照salary进行按照1-N的排名，相同salary并列且按照emp_no升序排列

select s1.emp_no,s1.salary,count(distinct s2.salary) rank

from salaries s1, salaries s2

where s1.salary <= s2.salary and s1.to_date = '9999-01-01' and s2.to_date = '9999-01-01'

group by s1.emp_no order by rank;

23.获取所有非manager员工当前的薪水情况，给出dept_no、emp_no以及salary ，当前表示to_date='9999-01-01'

SELECT de.dept_no,de.emp_no,sa.salary

FROM dept_emp de,salaries sa

WHERE de.emp_no=sa.emp_no AND de.to_date='9999-01-01' AND sa.to_date='9999-01-01' AND de.emp_no NOT IN(

    SELECT emp_no

    FROM dept_manager

)

24.获取员工其当前的薪水比其manager当前薪水还高的相关信息，当前表示to_date='9999-01-01',
结果第一列给出员工的emp_no，
第二列给出其manager的manager_no，
第三列给出该员工当前的薪水emp_salary,
第四列给该员工对应的manager当前的薪水manager_salary

SELECT s1.emp_no AS emp_no, s2.emp_no AS manager_no, s1.salary AS emp_salary, s2.salary AS manager_salary

FROM

(SELECT s.emp_no, de.dept_no, s.salary FROM salaries s INNER JOIN dept_emp de ON s.emp_no = de.emp_no AND s.to_date = '9999-01-01')

AS s1,

(SELECT s.emp_no, dm.dept_no, s.salary FROM salaries s INNER JOIN dept_manager dm ON s.emp_no = dm.emp_no AND s.to_date = '9999-01-01')

AS s2

WHERE s1.dept_no = s2.dept_no

AND s1.salary > s2.salary

25.汇总各个部门当前员工的title类型的分配数目，结果给出部门编号dept_no、dept_name、其当前员工所有的title以及该类型title对应的数目count

SELECT de.dept_no, dp.dept_name, t.title, COUNT(t.title) AS count

FROM titles AS t INNER JOIN dept_emp AS de

ON t.emp_no = de.emp_no AND de.to_date = '9999-01-01' AND t.to_date = '9999-01-01'

INNER JOIN departments AS dp

ON de.dept_no = dp.dept_no

GROUP BY de.dept_no, t.title

26.给出每个员工每年薪水涨幅超过5000的员工编号emp_no、薪水变更开始日期from_date以及薪水涨幅值salary_growth，并按照salary_growth逆序排列。

提示：在sqlite中获取datetime时间对应的年份函数为strftime('%Y', to_date)

SELECT s2.emp_no, s2.from_date, (s2.salary - s1.salary) AS salary_growth

 FROM salaries AS s1, salaries AS s2

 WHERE s1.emp_no = s2.emp_no

 AND s2.salary - s1.salary > 5000

 AND strftime('%Y',s2.to_date) - strftime('%Y',s1.to_date) = 1

 ORDER BY salary_growth DESC;

27.查找描述信息中包括robot的电影对应的分类名称以及电影数目，而且还需要该分类对应电影数量>=5部

select c.name,count(fcc.film_id)as num

from film_category fc

inner join category c on c.category_id=fc.category_id

inner join (select * from film f where f.description like '%robot%') as f on f.film_id=fc.film_id

inner join (select *,count (fc.film_id)as num from film_category fc group by category_id having num>=5)as fcc on fc.category_id=fcc.category_id

28.使用join查询方式找出没有分类的电影id以及名称

select f.film_id, f.title

from film as f

left join film_category as fc

on f.film_id = fc.film_id

where fc.category_id is null

29.使用子查询的方式找出属于Action分类的所有电影对应的title,description

 select title, description

from film

where film_id in (select film_id from film_category where category_id = (select category_id from category where name = 'Action'));

30.获取select * from employees对应的执行计划

explain select * from employees

31.将employees表的所有员工的last_name和first_name拼接起来作为Name，中间以一个空格区分

SELECT last_name||" "||first_name AS Name FROM employees

32.创建一个actor表，包含如下列信息

列表	类型	是否为NULL	含义
actor_id	smallint(5)	not null	主键id
first_name	varchar(45)	not null	名字
last_name	varchar(45)	not null	姓氏
last_update	timestamp	not null	最后更新时间，默认是系统的当前时间

CREATE TABLE actor

(

actor_id smallint(5) NOT NULL PRIMARY KEY,

first_name varchar(45) NOT NULL,

last_name varchar(45) NOT NULL,

last_update timestamp NOT NULL DEFAULT (datetime('now','localtime')) -- ,

-- PRIMARY KEY(actor_id)

)

33.对于表actor批量插入如下数据

CREATE TABLE IF NOT EXISTS actor (
actor_id smallint(5) NOT NULL PRIMARY KEY,
first_name varchar(45) NOT NULL,
last_name varchar(45) NOT NULL,
last_update timestamp NOT NULL DEFAULT (datetime('now','localtime')))

actor_id	first_name	last_name	last_update
1	PENELOPE	GUINESS	2006-02-15 12:34:33
2	NICK	WAHLBERG	2006-02-15 12:34:33

INSERT INTO actor

values (1,'PENELOPE','GUINESS','2006-02-15 12:34:33'),(2,'NICK','WAHLBERG','2006-02-15 12:34:33')

 

34.对于表actor批量插入如下数据,如果数据已经存在，请忽略，不使用replace操作
CREATE TABLE IF NOT EXISTS actor (
actor_id smallint(5) NOT NULL PRIMARY KEY,
first_name varchar(45) NOT NULL,
last_name varchar(45) NOT NULL,
last_update timestamp NOT NULL DEFAULT (datetime('now','localtime')))

actor_id	first_name	last_name	last_update
'3'	'ED'	'CHASE'	'2006-02-15 12:34:33'

INSERT or IGNORE INTO actor

VALUES ('3','ED','CHASE','2006-02-15 12:34:33');

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

35.对于如下表actor，其对应的数据为:

actor_id	first_name	last_name	last_update
1	PENELOPE	GUINESS	2006-02-15 12:34:33
2	NICK	WAHLBERG	2006-02-15 12:34:33

创建一个actor_name表，将actor表中的所有first_name以及last_name导入改表。 actor_name表结构如下：

列表	类型	是否为NULL	含义
first_name	varchar(45)	not null	名字
last_name	varchar(45)	not null	姓氏

create table actor_name(

    first_name varchar(45) not null,   

    last_name varchar(45) not null 

);

insert into actor_name select first_name,last_name from actor;

36.针对如下表actor结构创建索引：
CREATE TABLE IF NOT EXISTS actor (
actor_id smallint(5) NOT NULL PRIMARY KEY,
first_name varchar(45) NOT NULL,
last_name varchar(45) NOT NULL,
last_update timestamp NOT NULL DEFAULT (datetime('now','localtime')))
对first_name创建唯一索引uniq_idx_firstname，对last_name创建普通索引idx_lastname

create unique index uniq_idx_firstname on actor(first_name);

create index idx_lastname on actor(last_name);

37.针对actor表创建视图actor_name_view，只包含first_name以及last_name两列，并对这两列重新命名，first_name为first_name_v，last_name修改为last_name_v：
CREATE TABLE IF NOT EXISTS actor (
actor_id smallint(5) NOT NULL PRIMARY KEY,
first_name varchar(45) NOT NULL,
last_name varchar(45) NOT NULL,
last_update timestamp NOT NULL DEFAULT (datetime('now','localtime')))

create view actor_name_view (first_name_v,last_name_v) as

select first_name,last_name from actor

38.针对salaries表emp_no字段创建索引idx_emp_no，查询emp_no为10005, 使用强制索引。
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
create index idx_emp_no on salaries(emp_no);

select * from salaries

indexed by idx_emp_no

where emp_no = '10005'

39.存在actor表，包含如下列信息：
CREATE TABLE IF NOT EXISTS actor (
actor_id smallint(5) NOT NULL PRIMARY KEY,
first_name varchar(45) NOT NULL,
last_name varchar(45) NOT NULL,
last_update timestamp NOT NULL DEFAULT (datetime('now','localtime')));
现在在last_update后面新增加一列名字为create_date, 类型为datetime, NOT NULL，默认值为'0000 00:00:00'

ALTER TABLE actor ADD COLUMN create_date datetime NOT NULL DEFAULT '0000-00-00 00:00:00';

40.构造一个触发器audit_log，在向employees_test表中插入一条数据的时候，触发插入相关的数据到audit中。
CREATE TABLE employees_test(
ID INT PRIMARY KEY NOT NULL,
NAME TEXT NOT NULL,
AGE INT NOT NULL,
ADDRESS CHAR(50),
SALARY REAL
);
CREATE TABLE audit(
EMP_no INT NOT NULL,
NAME TEXT NOT NULL
);

create trigger audit_log after insert on employees_test

begin

     insert into audit values(new.id,new.name);

end;

41.删除emp_no重复的记录，只保留最小的id对应的记录。
CREATE TABLE IF NOT EXISTS titles_test (
id int(11) not null primary key,
emp_no int(11) NOT NULL,
title varchar(50) NOT NULL,
from_date date NOT NULL,
to_date date DEFAULT NULL);

insert into titles_test values ('1', '10001', 'Senior Engineer', '1986-06-26', '9999-01-01'),
('2', '10002', 'Staff', '1996-08-03', '9999-01-01'),
('3', '10003', 'Senior Engineer', '1995-12-03', '9999-01-01'),
('4', '10004', 'Senior Engineer', '1995-12-03', '9999-01-01'),
('5', '10001', 'Senior Engineer', '1986-06-26', '9999-01-01'),
('6', '10002', 'Staff', '1996-08-03', '9999-01-01'),
('7', '10003', 'Senior Engineer', '1995-12-03', '9999-01-01');

delete from titles_test

where id not in (select min(id) from titles_test)

42.将所有to_date为9999-01-01的全部更新为NULL,且 from_date更新为2001-01-01。
CREATE TABLE IF NOT EXISTS titles_test (
id int(11) not null primary key,
emp_no int(11) NOT NULL,
title varchar(50) NOT NULL,
from_date date NOT NULL,
to_date date DEFAULT NULL);

insert into titles_test values ('1', '10001', 'Senior Engineer', '1986-06-26', '9999-01-01'),
('2', '10002', 'Staff', '1996-08-03', '9999-01-01'),
('3', '10003', 'Senior Engineer', '1995-12-03', '9999-01-01'),
('4', '10004', 'Senior Engineer', '1995-12-03', '9999-01-01'),
('5', '10001', 'Senior Engineer', '1986-06-26', '9999-01-01'),
('6', '10002', 'Staff', '1996-08-03', '9999-01-01'),
('7', '10003', 'Senior Engineer', '1995-12-03', '9999-01-01');

update titles_test set to_date=null,from_date='2001-01-01' where to_date='9999-01-01';

43.将id=5以及emp_no=10001的行数据替换成id=5以及emp_no=10005,其他数据保持不变，使用replace实现。
CREATE TABLE IF NOT EXISTS titles_test (
id int(11) not null primary key,
emp_no int(11) NOT NULL,
title varchar(50) NOT NULL,
from_date date NOT NULL,
to_date date DEFAULT NULL);

insert into titles_test values ('1', '10001', 'Senior Engineer', '1986-06-26', '9999-01-01'),
('2', '10002', 'Staff', '1996-08-03', '9999-01-01'),
('3', '10003', 'Senior Engineer', '1995-12-03', '9999-01-01'),
('4', '10004', 'Senior Engineer', '1995-12-03', '9999-01-01'),
('5', '10001', 'Senior Engineer', '1986-06-26', '9999-01-01'),
('6', '10002', 'Staff', '1996-08-03', '9999-01-01'),
('7', '10003', 'Senior Engineer', '1995-12-03', '9999-01-01');

replace into titles_test values(5, 10005, 'Senior Engineer', '1986-06-26', '9999-01-01');

44.将titles_test表名修改为titles_2017。
CREATE TABLE IF NOT EXISTS titles_test (
id int(11) not null primary key,
emp_no int(11) NOT NULL,
title varchar(50) NOT NULL,
from_date date NOT NULL,
to_date date DEFAULT NULL);

insert into titles_test values ('1', '10001', 'Senior Engineer', '1986-06-26', '9999-01-01'),
('2', '10002', 'Staff', '1996-08-03', '9999-01-01'),
('3', '10003', 'Senior Engineer', '1995-12-03', '9999-01-01'),
('4', '10004', 'Senior Engineer', '1995-12-03', '9999-01-01'),
('5', '10001', 'Senior Engineer', '1986-06-26', '9999-01-01'),
('6', '10002', 'Staff', '1996-08-03', '9999-01-01'),
('7', '10003', 'Senior Engineer', '1995-12-03', '9999-01-01');

alter table titles_test rename to titles_2017

45.在audit表上创建外键约束，其emp_no对应employees_test表的主键id。
CREATE TABLE employees_test(
ID INT PRIMARY KEY NOT NULL,
NAME TEXT NOT NULL,
AGE INT NOT NULL,
ADDRESS CHAR(50),
SALARY REAL
);

CREATE TABLE audit(
EMP_no INT NOT NULL,
create_date datetime NOT NULL
);

DROP TABLE audit;

CREATE TABLE audit(

    EMP_no INT NOT NULL,

    create_date datetime NOT NULL,

    FOREIGN KEY(EMP_no) REFERENCES employees_test(ID));

 

 

46.存在如下的视图：
create view emp_v as select * from employees where emp_no >10005;
如何获取emp_v和employees有相同的数据？
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
输出格式:

emp_no	birth_date	first_name	last_name	gender	hire_date
10006	1953-04-20	Anneke	Preusig	F	1989-06-02
10007	1957-05-23	Tzvetan	Zielinski	F	1989-02-10
10008	1958-02-19	Saniya	Kalloufi	M	1994-09-15
10009	1952-04-19	Sumant	Peac	F	1985-02-18
10010	1963-06-01	Duangkaew	Piveteau	F	1989-08-24
10011	1953-11-07	Mary	Sluis	F	1990-01-22

select * from employees where emp_no >10005;

47.将所有获取奖金的员工当前的薪水增加10%。
create table emp_bonus(
emp_no int not null,
recevied datetime not null,
btype smallint not null);
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`from_date`));

update salaries set salary = salary*1.1

    where emp_no in (select emp_no from emp_bonus)

48.将所有获取奖金的员工当前的薪水增加10%。
create table emp_bonus(
emp_no int not null,
recevied datetime not null,
btype smallint not null);
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`from_date`));

update salaries set salary = salary*1.1

    where emp_no in (select emp_no from emp_bonus)

49.针对库中的所有表生成select count(*)对应的SQL语句
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
create table emp_bonus(
emp_no int not null,
recevied datetime not null,
btype smallint not null);
CREATE TABLE `dept_emp` (
`emp_no` int(11) NOT NULL,
`dept_no` char(4) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `dept_manager` (
`dept_no` char(4) NOT NULL,
`emp_no` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
输出格式:

cnts

select count(*) from employees;

select count(*) from departments;

select count(*) from dept_emp;

select count(*) from dept_manager;

select count(*) from salaries;

select count(*) from titles;

select count(*) from emp_bonus;

SELECT "select count(*) from " || name || ";" AS cnts

FROM sqlite_master WHERE type = 'table'

50.将employees表中的所有员工的last_name和first_name通过(')连接起来。
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
输出格式:

name

Facello'Georgi

Simmel'Bezalel

Bamford'Parto

Koblick'Chirstian

Maliniak'Kyoichi

Preusig'Anneke

Zielinski'Tzvetan

Kalloufi'Saniya

Peac'Sumant

Piveteau'Duangkaew

Sluis'Mary

select last_name || "'" || first_name as 'name'

from employees

 

51.查找字符串'10,A,B' 中逗号','出现的次数cnt。

select (length('10,A,B')-length(replace('10,A,B',',',''))) as cnt

52.获取Employees中的first_name，查询按照first_name最后两个字母，按照升序进行排列
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
输出格式：

first_name

Chirstian

Tzvetan

Bezalel

Duangkaew

Georgi

Kyoichi

Anneke

Sumant

Mary

Parto

Saniya

SELECT first_name FROM employees

order by substr(first_name,length(first_name)-1)

53.按照dept_no进行汇总，属于同一个部门的emp_no按照逗号进行连接，结果给出dept_no以及连接出的结果employees
CREATE TABLE `dept_emp` (
`emp_no` int(11) NOT NULL,
`dept_no` char(4) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
输出格式:

dept_no	employees
d001	10001,10002
d002	10006
d003	10005
d004	10003,10004
d005	10007,10008,10010
d006	10009,10010

SELECT dept_no, group_concat(emp_no) AS employees

FROM dept_emp GROUP BY dept_no;

54.查找排除当前最大、最小salary之后的员工的平均工资avg_salary。
CREATE TABLE `salaries` ( `emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
输出格式:

avg_salary

69462.5555555556

select avg(salary) as avg_salary from salaries

where to_date = '9999-01-01'

AND salary <> (select max(salary) from salaries)

AND salary <> (select min(salary) from salaries);

55.分页查询employees表，每5行一页，返回第2页的数据
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));

select * from employees limit 5,5

56.获取所有员工的emp_no、部门编号dept_no以及对应的bonus类型btype和recevied，没有分配具体的员工不显示
CREATE TABLE `dept_emp` ( `emp_no` int(11) NOT NULL,
`dept_no` char(4) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `dept_manager` (
`dept_no` char(4) NOT NULL,
`emp_no` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
输出格式:

e.emp_no	dept_no	btype	recevied
10001	d001	1	2010-01-01
10002	d001	2	2010-10-01
10003	d004	3	2011-12-03
10004	d004	1	2010-01-01
10005	d003
10006	d002
10007	d005
10008	d005
10009	d006
10010	d005
10010	d006

SELECT de.emp_no, de.dept_no, eb.btype, eb.recevied FROM dept_emp AS de LEFT JOIN emp_bonus AS eb

ON de.emp_no = eb.emp_no

 

57.使用含有关键字exists查找未分配具体部门的员工的所有信息。
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
CREATE TABLE `dept_emp` (
`emp_no` int(11) NOT NULL,
`dept_no` char(4) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
输出格式:

emp_no	birth_date	first_name	last_name	gender	hire_date
10011	1953-11-07	Mary	Sluis	F	1990-01-22

select*from employees where not exists (select emp_no from dept_emp where emp_no=employees.emp_no);

58.存在如下的视图：
create view emp_v as select * from employees where emp_no >10005;
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
获取employees中的行数据，且这些行也存在于emp_v中。注意不能使用intersect关键字。
输出格式:

emp_no	birth_date	first_name	last_name	gender	hire_date
10006	1953-04-20	Anneke	Preusig	F	1989-06-02
10007	1957-05-23	Tzvetan	Zielinski	F	1989-02-10
10008	1958-02-19	Saniya	Kalloufi	M	1994-09-15
10009	1952-04-19	Sumant	Peac	F	1985-02-18
10010	1963-06-01	Duangkaew	Piveteau	F	1989-08-24
10011	1953-11-07	Mary	Sluis	F	1990-01-22

select * from employees where emp_no >10005;

59.获取有奖金的员工相关信息。
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
CREATE TABLE `dept_emp` (
`emp_no` int(11) NOT NULL,
`dept_no` char(4) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
create table emp_bonus(
emp_no int not null,
recevied datetime not null,
btype smallint not null);
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL, PRIMARY KEY (`emp_no`,`from_date`));
给出emp_no、first_name、last_name、奖金类型btype、对应的当前薪水情况salary以及奖金金额bonus。 bonus类型btype为1其奖金为薪水salary的10%，btype为2其奖金为薪水的20%，其他类型均为薪水的30%。 当前薪水表示to_date='9999-01-01'
输出格式:

emp_no	first_name	last_name	btype	salary	bonus
10001	Georgi	Facello	1	88958	8895.8
10002	Bezalel	Simmel	2	72527	14505.4
10003	Parto	Bamford	3	43311	12993.3
10004	Chirstian	Koblick	1	74057	7405.7

 

 

 

 

 

 

select e.emp_no,e.first_name,e.last_name,b.btype,s.salary,

(case b.btype

when 1 then s.salary*0.1

when 2 then s.salary*0.2

else s.salary*0.3 end) as bonus

from employees e inner join emp_bonus b on e.emp_no=b.emp_no inner join salaries s on

s.emp_no=e.emp_no and s.to_date = '9999-01-01';

60.按照salary的累计和running_total，其中running_total为前两个员工的salary累计和，其他以此类推。 具体结果如下Demo展示。。
CREATE TABLE `salaries` ( `emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
输出格式:

emp_no	salary	running_total
10001	88958	88958
10002	72527	161485
10003	43311	204796
10004	74057	278853
10005	94692	373545
10006	43311	416856
10007	88070	504926
10009	95409	600335
10010	94409	694744
10011	25828	720572

select a.emp_no,a.salary,(select sum(s.salary) from  salaries as s where s.emp_no <=a.emp_no and s.to_date='9999-01-01')as running_total

from salaries as a where a.to_date='9999-01-01' order by a.emp_no;

61.对于employees表中，给出奇数行的first_name
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
输出格式:

first_name

Georgi

Chirstian

Anneke

Tzvetan

Saniya

Mary

SELECT E1.first_name

FROM employees E1

WHERE(SELECT COUNT(*) FROM employees E2 WHERE E1.first_name>=E2.first_name)%2=1;