【1】film表
| 字段 | 说明 |
|---|---|
| film_id | 电影id |
| title | 电影名称 |
| description | 电影描述信息 |
category表
| 字段 | 说明 |
|---|---|
| category_id | 电影分类id |
| name | 电影分类名称 |
| last_update | 电影分类最后更新时间 |
film_category表
| 字段 | 说明 |
|---|---|
| film_id | 电影id |
| category_id | 电影分类id |
| last_update | 电影id和分类id对应关系的最后更新时间 |
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
说明：cc是一个虚表，因为虚表cc是按category_id进行了分组，这意味着每个category_id分类下只返回一条数据，所以无法替代film_category表来做为连接表
select c.name,count(c.name) from
(select category_id,count(film_id) as s from film_category group by category_id having s>=5) as cc,
category c,film f,film_category fc where
f.film_id=fc.film_id and c.category_id=fc.category_id and c.category_id=cc.category_id and
f.description like '%robot%';
【2】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));
SELECT e1.first_name FROM
(SELECT e2.first_name,
(SELECT COUNT(*) FROM employees AS e3
WHERE e3.first_name <= e2.first_name)
AS rowid FROM employees AS e2) AS e1
WHERE e1.rowid % 2 = 1
【3】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’
select e.emp_no,e.first_name,e.last_name,eb.btype,s.salary,
(case eb.btype
when 1 then s.salary*0.1
when 2 then s.salary*0.2
else s.salary*0.3 end)as bonus
from salaries s,emp_bonus eb,employees e
where s.to_date='9999-01-01' and s.emp_no=e.emp_no and s.emp_no=eb.emp_no;
【4】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));
select s1.emp_no,s1.from_date,(s1.salary-s2.salary) as salary_growth
from salaries s1,salaries s2 where s1.emp_no=s2.emp_no
and (strftime('%Y',s1.to_date)-strftime('%Y',s2.to_date)=1)
and (s1.salary-s2.salary)>5000
order by salary_growth desc;
【5】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));
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));
select e.emp_no,max(s.salary)as salary,e.last_name,e.first_name from employees e,salaries s
where e.emp_no=s.emp_no and s.to_date='9999-01-01' and
s.salary<(select max(salary) from salaries where to_date='9999-01-01');
【6】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));
select s1.emp_no,s1.salary,
(select count(distinct salary) from salaries where salary>=s1.salary and to_date='9999-01-01')as rank
from salaries s1 where s1.to_date='9999-01-01'
order by rank,s1.emp_no;
【7】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));
提示：
limit i,n
i：为查询结果的索引值(默认从0开始)，当i=0时可省略i
n：为查询结果返回的数量
select emp_no,salary from salaries
where salary=
(select distinct salary from salaries order by salary desc limit 1,1);
【8】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));
select a.emp_no,count(*) t from salaries a inner join salaries b
on a.emp_no=b.emp_no and a.to_date = b.from_date
where a.salary < b.salary
group by a.emp_no
having t>15;
【9】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));
提示：
LIMIT m,n : 表示从第m+1条开始，取n条数据；
LIMIT n ： 表示从第0条开始，取n条数据，是limit(0,n)的缩写。
（1）首先需要加distinct去重。
假设 5-23（入职最晚日期）入职的有a,b,c 3人；5-22（入职第二晚日期）入职的有d,e 2人；5-21(入职倒数第三晚)入职的有f,g,h 3人；5-21前入职的若干…若不加distinct去重，那么按照日期倒序，limit 2,1（从倒数第2行开始，取一条数据）的查询结果为 5-23；加了distinct去重，会按入职日期进行分组，多个相同入职日期会分为一组，这样limit 2,1的结果即为 5-21。
（2）外层的where条件中根据子查询查出的倒数第三晚入职的日期，就能查询出符合条件的员工信息。
select * from employees
where hire_date = (
select distinct hire_date from employees order by hire_date desc limit 2,1
);
【10】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));
select s1.emp_no,s1.salary,sum(s2.salary)as running_total from salaries s2,salaries s1
where s2.emp_no<=s1.emp_no and s2.to_date='9999-01-01' and s1.to_date='9999-01-01'
group by s1.emp_no;
SELECT s1.emp_no, s1.salary,
(SELECT SUM(s2.salary) FROM salaries AS s2
WHERE s2.emp_no <= s1.emp_no AND s2.to_date = '9999-01-01') AS running_total
FROM salaries AS s1 WHERE s1.to_date = '9999-01-01' ORDER BY s1.emp_no;
来源：CSDN
作者：K. Bob
链接：https://blog.csdn.net/ThreeAspects/article/details/103764728