数据库练习SQL
1. 查找最晚入职员工的所有信息
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`));
【解答】select * from employees order by hire_date desc limit 1;
2. 查找入职员工时间排名倒数第三的员工所有信息
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`));
【解答】select emp_no,birth_date,first_name,last_name,gender,hire_date
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
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`));
【解析】
select salaries.*,dept.dept_no
from salaries salaries left join dept_manager dept on
salaries.emp_no=dept.emp_no
where salaries.to_date='9999-01-01' and dept.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
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`));
【解析】
select emp.last_name,emp.first_name,dep.dept_no
from employees emp ,dept_emp dep
Where emp.emp_no = dep.emp_no
5.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
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`));
【解析】
select t.emp_no,p.salary
from employees t inner join salaries p
on t.emp_no=p.emp_no and t.hire_date=p.from_date order by p.emp_no desc
6.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t
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`));
【解析】 select emp_no,count(emp_no) as t from salaries group by emp_no having t>15
7.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
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`));
【解析】
Select distinct(salary) from salaries where o_date='9999-01-01'
8.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'
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`));
【解析】
select m.dept_no,m.emp_no,s.salary
from dept_manager m inner join salaries s
on m.emp_no=s.emp_no and m.to_date=s.to_date and s.to_date='9999-01-01'
9.获取所有非manager的员工emp_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`));
【解析】select emp_no from employees not in (
Select emp_no from dept_manager )
10.获取所有员工当前的manager，如果当前的manager是自己的话结果不显示，当前表示to_date='9999-01-01'。
结果第一列给出当前员工的emp_no,第二列给出其manager对应的manager_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`));
【解析】
SELECT de.emp_no,dm.emp_no AS manager_no
FROM dept_manager AS dm,dept_emp AS de
WHERE de.emp_no <> dm.emp_no AND de.dept_no = dm.dept_no
AND dm.to_date='9999-01-01';
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
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`));
【解析】
A select a.dept_no,a.emp_no,max(a.salary) from(
select t.emp_no,t.dept_no,s.salary
from dept_emp t,salaries s
where t.emp_no==s.emp_no and t.to_date = '9999-01-01' AND s.to_date = '9999-01-01') a group by a.dept_no
B select d.dept_no,d.emp_no,s.salary
From dept_emp d,(select emp_no ,salary from salaries order by salary desc limit 1) s
Where d.emp_no=s.emp_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。
CREATE TABLE IF NOT EXISTS "titles" (
`emp_no` int(11) NOT NULL,
`title` varchar(50) NOT NULL,
`from_date` date NOT NULL,
`to_date` date DEFAULT NULL);
【解析】select title, count(distinct emp_no) as t from titles group by titles having t>=2
13.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
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`));
【解析】select * from employees t where t.last_name!=’Mary’ and emp_no%2==1
order by hire_date desc
14.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。
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`));
【解析】
A. select a.title,avg(a.salary) as avg
from(
select t.emp_no,s.salary,t.title
from titles t,salaries s
where t.emp_no=s.emp_no and t.to_date = '9999-01-01' AND s.to_date = '9999-01-01') a group by a.title
15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
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`));
【解析】select emp_no, salary
from salaries where to_date=’9999-01-01’ order by salary limit 1,1
关于关键词limit的描述：limit是mysql的语法select * from table limit m,n其中m是指记录开始的index，从0开始，表示第一条记录n是指从第m+1条开始，取n条;
16.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
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`));
【解析】
A. select s.emp_no,max(s.salary),e.last_name,e.first_name
from employees e,salaries s
where s.emp_no=e.emp_no and s.to_date="9999-01-01"
and s.salary !=(select max(salary) from salaries) and e.emp_no=s.emp_no;
B. select e.emp_no,a.salary
from employee e,(
select emp_no, Max(salary) as salary from salaries
where salary !=(
Select Max(salary) from salaries) ) a
Where e.emp_no=a.emp_no and to_date="9999-01-01";
17.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
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`));
【解析】
select e.last_name,e.first_name,de.dept_name from employees e
left join dept_emp t on t.emp_no=e.emp_no
left join departments de on de.dept_no=t.dept_no
18.查找员工编号emp_now为10001其自入职以来的薪水salary涨幅值growth
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`));
【解析】
Select emp_no,(Max(salary)-Min(salary)) as growth
from salaries where emp_no=’10001’
19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_noy以及其对应的薪水涨幅growth，并按照growth进行升序
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`));
【解析】
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, --现在的工资
(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 --原始的工资
WHERE sCurrent.emp_no = sStart.emp_no
ORDER BY growth
20.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
CREATE TABLE `departments` (
`dept_no` char(4) NOT NULL,
`dept_name` varchar(40) NOT NULL,
PRIMARY KEY (`dept_no`));
【解析】
SELECT
de.dept_no, dp.dept_name, COUNT(s.salary) AS sum
FROM ( dept_emp AS de INNER JOIN salaries AS s ON de.emp_no = s.emp_no)
INNER JOIN departments AS dp ON de.dept_no = dp.dept_no
GROUP BY de.dept_no
21.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
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`));
【解析】
SELECT s1.emp_no, s1.salary, COUNT(DISTINCT s2.salary) AS rank
FROM salaries AS s1, salaries AS s2
WHERE s1.to_date = '9999-01-01' AND s2.to_date = '9999-01-01' AND s1.salary <= s2.salary
GROUP BY s1.emp_no
ORDER BY s1.salary DESC, s1.emp_no ASC
22.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'
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 `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`));
【解析】
Select e.dept_no,e.emp_no,s.salary
From dept_emp e
Inner join salaries s
on e.emp_no=s.emp_no and s.todate=’9999-01-01’
Where e.emp_no not in
(select distinct(emp_no) from dept_manager where todate=’9999-01-01’)
23. 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
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`));
【解析】
select w.emp_no as emp_no,p.emp_no as manager_no,w.salary as emp_salary,p.salary as manager_salary
from
(select d.emp_no,s.salary,d.dept_no from dept_emp d inner join salaries s on d.emp_no=s.emp_no and s.to_date='9999-01-01') w,
(select m.emp_no,s.salary,m.dept_no from dept_manager m inner join salaries s on m.emp_no=s.emp_no and s.to_date='9999-01-01') p
where w.dept_no=p.dept_no and w.salary > p.salary
24. 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
CREATE TABLE `departments` (
`dept_no` char(4) NOT NULL,
`dept_name` varchar(40) NOT NULL,
PRIMARY KEY (`dept_no`));
CREATE TABLE IF NOT EXISTS "titles" (
`emp_no` int(11) NOT NULL,
`title` varchar(50) NOT NULL,
`from_date` date NOT NULL,
`to_date` date DEFAULT NULL);
【解析】select a.dept_no,
(select dept_name from departments where dept_no = a.dept_no) as 'dept_name',
b.title,
count(*) as 'count'
from dept_emp as a inner join titles as b on a.emp_no = b.emp_no--统计部门的tile的各个中类型的个数
where a.to_date = '9999-01-01'
and b.to_date = '9999-01-01'
group by a.dept_no, b.title--根据部门代码和title种类进行分类
order by a.dept_no a.dept_no
25.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)
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`));
【解析】：
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 salary_growth > 5000
AND (strftime("%Y",s2.to_date) - strftime("%Y",s1.to_date) = 1
OR strftime("%Y",s2.from_date) - strftime("%Y",s1.from_date) = 1 )
ORDER BY salary_growth DESC
26.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
CREATE TABLE IF NOT EXISTS film (
film_id smallint(5) NOT NULL DEFAULT '0',
title varchar(255) NOT NULL,
description text,
PRIMARY KEY (film_id));
CREATE TABLE category (
category_id tinyint(3) NOT NULL ,
name varchar(25) NOT NULL, `last_update` timestamp,
PRIMARY KEY ( category_id ));
film_category表
CREATE TABLE film_category (
film_id smallint(5) NOT NULL,
category_id tinyint(3) NOT NULL, `last_update` timestamp);
【解析】SELECT c.name, COUNT(fc.film_id) FROM
(select category_id, COUNT(film_id) AS category_num FROM film_category GROUP BY category_id HAVING count(film_id)>=5) AS cc,
film AS f, film_category AS fc, category AS c
WHERE f.description LIKE '%robot%'
AND f.film_id = fc.film_id
AND c.category_id = fc.category_id
AND c.category_id=cc.category_id
27.使用join查询方式找出没有分类的电影id以及名称
【film表】
CREATE TABLE IF NOT EXISTS film (
film_id smallint(5) NOT NULL DEFAULT '0',
title varchar(255) NOT NULL,
description text,
PRIMARY KEY (film_id));
【category表】
CREATE TABLE category (
category_id tinyint(3) NOT NULL ,
name varchar(25) NOT NULL, `last_update` timestamp,
PRIMARY KEY ( category_id ));
【film_category表】
CREATE TABLE film_category (
film_id smallint(5) NOT NULL,
category_id tinyint(3) NOT NULL, `last_update` timestamp);
【解析】
Select a.id,a.title from(
Select f.id,f.titile,fc.category_id
From film f left join film_category fc on f.film_id=cf.film_id ) a
Where a.category_id is null;
28.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
CREATE TABLE IF NOT EXISTS film (
film_id smallint(5) NOT NULL DEFAULT '0',
title varchar(255) NOT NULL,
description text,
PRIMARY KEY (film_id));
CREATE TABLE category (
category_id tinyint(3) NOT NULL ,
name varchar(25) NOT NULL, `last_update` timestamp,
PRIMARY KEY ( category_id ));
CREATE TABLE film_category (
film_id smallint(5) NOT NULL,
category_id tinyint(3) NOT NULL, `last_update` timestamp);
【解析】
A.
select f.title,f.description
from film as f inner join film_category as fc on f.film_id = fc.film_id
inner join category as c on c.category_id = fc.category_id
where c.name = 'Action';
B.
select title, description
from film where film_id in (select fc.film_id
from film_category fc
where fc.category_id in
(select c.category_id
from category c
where c.name = 'Action'))
29.获取select * from employees对应的执行计划
【解析】
EXPLAIN SELECT * FROM employees
30.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
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`));
【解析】SELECT last_name||" "||first_name AS Name FROM employees
31.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
【解析】
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'))
)
32.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 |
【解析】
Insert into actor(actor_id,first_name,last_name,last_update)
Values (1, 'PENELOPE', 'GUINESS', '2006-02-15 12:34:33'),
(2, 'NICK', 'WAHLBERG', '2006-02-15 12:34:33');
33.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
【解析】
insert or ignore into actor values ('3','ED','CHASE','2006-02-15 12:34:33')
-- insert or replace：如果不存在就插入，存在就更新
-- insert or ignore： 如果不存在就插入，存在就忽略
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
【解析】create table actor_name as select first_name,last_name from actor
34.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
【解析】create unique index uniq_idx_firstname on actor(first_name);
create index idx_lastname on actor(last_name);
35.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，fist_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')))
【解析】
CREATE VIEW actor_name_view AS
select first_name as fist_name_v,last_name as last_name_v from actor
36. 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);
【解析】SELECT * FROM salaries INDEXED BY idx_emp_no WHERE emp_no = 10005
【关于索引的描述】
1. 创建索引：create 【unique/】 index 索引名字 on表明（列名）
2. 对于一张表，如果表中的某一列设置了主键，则会在该列上创建一个索引，主索引unique index
3. 查询的时候走索引，可以提高查询的速度，使用强制索引的方法
Select * from 表名 indexed by 索引名 where 需要强制走索引的列名 +条件
4. 索引，如果有索引，先从索引走，占用空间比较大，索引是加在字段上的，orcale自动回创建索引，orcale回自动创建一个索引，会给款卫衣的约束。如果where字段使用了函数，则不走函数。
5. Drop index 索引名字
6. 创建组合索引，crate index 索引名字 on 表明（列名1，列名2）
37.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'
【解析】
Alter table actor add create_date datetime default ‘0000-00-00 00:00:00’ not null
37. 构造一个触发器audit_log，在向employees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
);
【解析】
create trigger audit_log
after insert on employees_test
for each row
Begin
insert into audit(EMP_no,NAME) values(new.ID,new.NAME);
END;
38.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);
【解析】
DELETE FROM titles_test WHERE id NOT IN (
SELECT MIN(id) FROM titles_test GROUP BY emp_no)
39.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);
【解析】
UPDATE titles_test SET to_date = NULL, from_date = '2001-01-01'
WHERE to_date = '9999-01-01';
40.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);
【解析】
Update titles_test set emp_no=replace(emp_no,’10001’,’10005’) where id=5;
41.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);
【解析】
Alter table titles_tes rename to titles_2017;
42.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
);
【解析】
A. DROP TABLE audit;
CREATE TABLE audit(
EMP_no INT NOT NULL,
create_date datetime NOT NULL,
FOREIGN KEY(EMP_no) REFERENCES employees_test(ID));
B. alter table audit add foreign key(EMP_no) references employees_test(ID);
43.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？
【解析】
Select * from employees where emp_no >10005;
44.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`));
【解析】update emp_bonus set salary=salary*1.1
45.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 `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;
【解析】SELECT "select count(*) from " || name || ";" AS cnts
FROM sqlite_master WHERE type = 'table'
46. 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`));
【解析】
Select last_name || “’” || first_name from employee;
47. 查找字符串'10,A,B' 中逗号','出现的次数cnt
【解析】
SELECT (length("10,A,B")-length(replace("10,A,B",",","")))/length(",") AS cnt from dual;
48.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
【解析】
A. SELECT first_name FROM employees ORDER BY substr(first_name,-2)
B. select t.first_name FROM employees t
order by substr(t.first_name,length(t.first_name)-1,length(t.first_name));
【关于subStr函数的描述】
49.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`));
【解析】SELECT dept_no, group_concat(emp_no) as employees
from dept_emp group by dept_no;[此函数MYSQL]
【关于聚合函数的相关描述】
(select t.nsrmc ,(wm_concat( distinct(t.qxqybz))) as a from tax_source_by_details t where t.nd=2018 group by t.nsrmc)
50.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
【解析】
A. Select avg(salary) as avg_salary
From salaries where salary !=(select min(salary) from salaries)
And salary !=(select max(salary) from salaries)
And to_date = '9999-01-01' ;
B SELECT AVG(salary) AS avg_salary FROM salaries
WHERE to_date = '9999-01-01'
AND salary NOT IN (SELECT MAX(salary) FROM salaries)
AND salary NOT IN (SELECT MIN(salary) FROM salaries)
51.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`));
【解析】
方法一：利用 LIMIT 和 OFFSET 关键字。LIMIT 后的数字代表返回几条记录，OFFSET 后的数字代表从第几条记录开始返回（第一条记录序号为0），也可理解为跳过多少条记录后开始返回。
SELECT * FROM employees LIMIT 5 OFFSET 5
方法二：只利用 LIMIT 关键字。注意：在 LIMIT X,Y 中，Y代表返回几条记录，X代表从第几条记录开始返回（第一条记录序号为0），切勿记反。
SELECT * FROM employees LIMIT 5,5
方法三：常见的分页语句
Select rownum ,t.* from
( select * from 表名字 );此时，原表上面就会存在行号
52.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`));
【解析】
Select * from employees t not exists (
Select de.emp_no from dept_emp de where de.emp_no=t.emp_no);
53.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`));
【解析】
A. Select first_name
From employee e1
where ( select count(*) from employees as e2
where e2.first_name <=e1.first_name ) %2 =1
B.select a.first_name
from (select rownum as w, t.* from (select t.* from employee t order by t.frist_name) t) a
where mod(a.w , 2 )= 1;