查找最晚入职员工的所有信息
select * from employees order by hire_date desc limit 0,1
查找入职员工时间排名倒数第三的员工所有信息
select * from employees order by hire_date desc limit 2,1
查找当前薪水详情以及部门编号dept_no
select salaries.*,dept_manager.dept_no from
salaries, dept_manager
where salaries.emp_no = dept_manager.emp_no
and salaries.to_date = "9999-01-01"
and dept_manager.to_date = "9999-01-01";
查找所有已经分配部门的员工的last_name和first_name
select employees.last_name, employees.first_name, dept_emp.dept_no
from employees, dept_emp
where employees.emp_no = dept_emp.emp_no;
查找所有员工的last_name和first_name以及对应部门编号dept_no
select employees.last_name, employees.first_name, dept_emp.dept_no
from employees left join dept_emp
on dept_emp.emp_no = employees.emp_no;
查找所有员工入职时候的薪水情况
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
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
select emp_no, count(*) as t
from salaries
group by emp_no
having t>15
找出所有员工当前薪水salary情况
select distinct salary from salaries
where to_date = '9999-01-01'
order by salary DESC
获取所有部门当前manager的当前薪水情况
select dept_manager.dept_no, dept_manager.emp_no, salaries.salary
from dept_manager, salaries
where dept_manager.emp_no = salaries.emp_no
and salaries.to_date = '9999-01-01'
and dept_manager.to_date = '9999-01-01'
order by dept_manager.emp_no;
获取所有非manager的员工emp_no
select emp_no from employees
where emp_no not in(select emp_no from dept_manager)
获取所有员工当前的manager
select dept_emp.emp_no, dept_manager.emp_no
from dept_emp, dept_manager where
dept_emp.dept_no = dept_manager.dept_no and
dept_emp.emp_no != dept_manager.emp_no and
dept_emp.to_date = '9999-01-01' and
dept_manager.to_date = '9999-01-01' ;
获取所有部门中当前员工薪水最高的相关信息
select d.dept_no, s.emp_no, max(s.salary) as salary
from salaries as s inner join dept_emp as d
on d.emp_no = s.emp_no
where d.to_date = '9999-01-01' and s.to_date = '9999-01-01'
group by d.dept_no
从titles表获取按照title进行分组
select title,count(*) as t from titles
group by title
having t>1;
从titles表获取按照title进行分组，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>1;
查找employees表
select * from employees where
emp_no%2 == 1 and
last_name != 'Mary'
order by hire_date DESC;
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
select title,avg(salary) from
salaries as s inner join titles as t on s.emp_no = t.emp_no
where s.to_date = '9999-01-01' and t.to_date = '9999-01-01'
group by t.title;
获取当前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
order by salary DESC
limit 1,1;
获取当前薪水第二多的员工的emp_no以及其对应的薪水salary，不使用order by
select e.emp_no,max(s.salary),e.last_name,e.first_name from
employees as e inner join salaries as s on e.emp_no = s.emp_no
where salary != (select max(salary) from salaries where to_date = '9999-01-01');
查找所有员工的last_name和first_name以及对应的dept_name
select e.last_name, e.first_name, d.dept_name from
(employees as e left join dept_emp on e.emp_no = dept_emp.emp_no)
left join departments as d on d.dept_no = dept_emp.dept_no;
查找员工编号emp_now为10001其自入职以来的薪水salary涨幅值
select max(salary)-min(salary) from salaries where emp_no = '10001';
查找所有员工自入职以来的薪水涨幅情况
select sCurrent.emp_no, (sCurrent.salary - sStart.salary) as growth
from (select s.emp_no, s.salary from employees as e left join salaries as s on e.emp_no = s.emp_no where s.to_date = '9999-01-01') as sCurrent
inner join (select s.emp_no, s.salary from employees as e left join salaries as s on e.emp_no = s.emp_no where s.from_date = e.hire_date) as sStart
on sCurrent.emp_no = sStart.emp_no
order by growth
统计各个部门对应员工涨幅的次数总和
select de.dept_no, dp.dept_name, count(sa.salary) as sum
from (dept_emp as de inner join salaries as sa on de.emp_no = sa.emp_no)
inner join departments as dp on de.dept_no = dp.dept_no
group by de.dept_no
对所有员工的薪水按照salary进行按照1-N的排名
select s1.emp_no, s1.salary, count(distinct s2.salary)
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
获取所有非manager员工当前的薪水情况
select de.dept_no, em.emp_no, sa.salary from
(employees em inner join salaries sa on em.emp_no = sa.emp_no and sa.to_date = '9999-01-01')
inner join dept_emp de on de.emp_no = sa.emp_no where
em.emp_no not in (select emp_no from dept_manager where to_date = '9999-01-01');
获取员工其当前的薪水比其manager当前薪水还高的相关信息
select de.emp_no, dm.emp_no, s1.salary, s2.salary from
dept_emp de, dept_manager dm, salaries s1, salaries s2 where
de.dept_no = dm.dept_no and
de.emp_no = s1.emp_no and s1.to_date = '9999-01-01' and
dm.emp_no = s2.emp_no and s2.to_date = '9999-01-01' and
s1.salary > s2.salary;
汇总各个部门当前员工的title类型的分配数目
select de.dept_no, de.dept_name, ti.title, count(ti.title) as cnt
from titles as ti inner join dept_emp as d
on ti.emp_no = d.emp_no and ti.to_date = '9999-01-01' and d.to_date = '9999-01-01'
inner join departments as de on de.dept_no = d.dept_no
group by de.dept_no, ti.title;
给出每个员工每年薪水涨幅超过5000的员工编号emp_no
select s1.emp_no, s1.from_date, (s1.salary - s2.salary) as salary_growth from
salaries as s1, salaries as s2 where
s1.emp_no = s2.emp_no and
s1.salary - s2.salary > 5000 and
( strftime('%Y', s1.to_date) - strftime("%Y",s2.to_date) = 1 or
strftime("%Y", s1.from_date) - strftime("%Y",s2.from_date) = 1 )
order by salary_growth desc
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
select c.name, count(c.name) as number from
film as f inner join film_category as fc on f.film_id = fc.film_id
inner join category as c on fc.category_id = c.category_id
where f.description like '%robot%'
group by c.name
having number > 1;
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;
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 f.title, f.description from
film as f, film_category as fc, category c where
f.film_id = fc.film_id and
fc.category_id = c.category_id and
c.name = 'Action';
获取select * from employees对应的执行计划
explain select * from employees
将employees表的所有员工的last_name和first_name拼接起来作为Name
select (last_name || ' ' || first_name) as Name from employees
创建一个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'))
批量插入数据
insert into actor values (1, 'PENELOPE', 'GUINESS', '2006-02-15 12:34:33'),(2, 'NICK', 'WAHLBERG', '2006-02-15 12:34:33');
)
批量插入数据，不使用replace操作
insert or ignore into actor values ('3', 'ED', 'CHASE', '2006-02-15 12:34:33');
创建一个actor_name表
create table actor_name as select first_name, last_name from actor;
对first_name创建唯一索引uniq_idx_firstname
create unique index uniq_idx_firstname on actor(first_name);
create index idx_lastname on actor(last_name);
针对actor表创建视图actor_name_view
create view actor_name_view as select first_name as first_name_v, last_name as last_name_v from actor;
针对上面的salaries表emp_no字段创建索引idx_emp_no
select * from salaries indexed by idx_emp_no where emp_no = 10005;
在last_update后面新增加一列名字为create_date
alter table actor add column create_date datetime not null default('0000-00-00 00:00:00');
构造一个触发器audit_log
create trigger audit_log after insert on employees_test begin insert into audit values(new.id, new.name);
end;
删除emp_no重复的记录，只保留最小的id对应的记录。
delete from titles_test where id not in(select min(id) from titles_test group by emp_no);
将所有to_date为9999-01-01的全部更新为NULL
update titles_test set to_date = null, from_date = '2001-01-01' where to_date = '9999-01-01';
将id=5以及emp_no=10001的行数据替换成id=5以及emp_no=10005
update titles_test set emp_no = replace(emp_no,10001,10005) where id = 5;
将titles_test表名修改为titles_2017
alter table titles_test rename to titles_2017;
在audit表上创建外键约束，其emp_no对应employees_test表的主键id
drop table audit;
create table audit(
EMP_no int not null,
create_date datetime not null,
foreign key(EMP_no) references employees_test(ID));
如何获取emp_v和employees有相同的数据no
select * from emp_v intersect select * from employees;
将所有获取奖金的员工当前的薪水增加10%
update salaries set salary = salary*1.1 where emp_no in
(select s.emp_no from emp_bonus as e, salaries as s where
s.to_date = '9999-01-01' and s.emp_no = e.emp_no);
针对库中的所有表生成select count(*)对应的SQL语句
select "select count(*) from " || name || ";" as cnts from
sqlite_master where type = 'table';
将employees表中的所有员工的last_name和first_name通过(‘)连接起
select last_name || "'" || first_name as name from employees;
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;
获取Employees中的first_name
select first_name from employees order by substr(first_name, -2);
按照dept_no进行汇总
select dept_no, group_concat(emp_no) as employees from dept_emp group by dept_no;
查找排除当前最大、最小salary之后的员工的平均工资avg_salary
select avg(salary) from salaries where
to_date = '9999-01-01' and
salary not in(select max(salary) from salaries) and (select min(salary) from salaries);
分页查询employees表，每5行一页，返回第2页的数据
select * from employees limit 5,5
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。
select * from employees where emp_no not in (select emp_no from dept_emp)
//不知道怎么使用exist。。
获取employees中的行数据，且这些行也存在于emp_v中
select * from employees where emp_no > 10005;
获取有奖金的员工相关信息。
select em.emp_no, em.first_name, em.last_name, eb.btype, sa.salary,
(case when eb.btype<=2 then eb.btype*0.1*sa.salary when eb.btype>=3 then 0.3*sa.salary end ) as bonus
from employees as em inner join salaries as sa on em.emp_no=sa.emp_no
inner join emp_bonus as eb on em.emp_no=eb.emp_no and sa.to_date='9999-01-01';
统计salary的累计和running_total
select s2.emp_no, s2.salary, (select sum(s3.salary) from salaries s3 where s2.emp_no >= s3.emp_no and s3.to_date = '9999-01-01') as runing_total
from salaries s2 where s2.to_date = '9999-01-01'
order by s2.emp_no;
对于employees表中，给出奇数行的first_name
select e1.first_name from employees as e1 where
(select count(*) from employees as e2 where e1.first_name >= e2.first_name)%2 = 1;
