相关操作
查找最晚入职员工的所有信息 SELECT * FROM employees ORDER BY hire_date desc limit 0,1;/*limit m, n 表示从m+1开始去n条数据*/ 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 s.*, d.dept_no from salaries s,dept_manager d where s.to_date = '9999-01-01' and d.to_date = '9999-01-01'and s.emp_no = d.emp_no; 查找所有已经分配部门的员工的last_name和first_name select e.last_name,e.first_name,d.dept_no from employees e,dept_emp d where e.emp_no = d.emp_no; 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 e.last_name,e.first_name,d.dept_no from employees e left join dept_emp d on e.emp_no = d.emp_no; 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 e.emp_no,s.salary from employees as e , salaries as s where s.emp_no=e.emp_no and s.from_date=e.hire_date order by s.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(*) t from salaries group by emp_no having t > 15; 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; 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 d.dept_no,d.emp_no,s.salary from salaries as s, dept_manager as d where d.emp_no = s.emp_no and d.to_date = '9999-01-01' and s.to_date = '9999-01-01'; 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，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。 select de.emp_no,dm.emp_no from dept_emp de, dept_manager dm where de.dept_no = dm.dept_no and de.emp_no not in (select emp_no from dept_manager) and de.to_date = '9999-01-01' and dm.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，给出dept_no, emp_no以及其对应的salary SELECT d.dept_no, d.emp_no, MAX(s.salary) AS salary FROM salaries AS s, dept_emp AS d WHERE s.emp_no = d.emp_no AND 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，每组个数大于等于2，给出title以及对应的数目t。 select title,count(*) t from titles group by title having t >= 2; 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) t from titles group by title having t >= 2; 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 last_name != 'Mary' and round(emp_no / 2.0) != emp_no / 2.0 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。结果给出title以及平均工资avg。 select t.title, avg(s.salary) from salaries s, titles t where t.to_date = '9999-01-01' and s.to_date = '9999-01-01' and t.emp_no = s.emp_no group by t.title; 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 where to_date = '9999-01-01' order by salary desc limit 1,1; 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), e.last_name, e.first_name from employees e, salaries s where s.salary < (select MAX(salary) from salaries) and e.emp_no = s.emp_no; 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,d.dept_name from employees e left join dept_emp de on e.emp_no = de.emp_no left join departments d on de.dept_no = d.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growth select (MAX(salary) - MIN(salary))growth from salaries where emp_no = 10001; 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 select t1.emp_no,t1.salary-t2.salary as growth from (select s.emp_no,salary from salaries as s,employees as e where s.emp_no=e.emp_no and s.to_date='9999-01-01')as t1, (select s.emp_no,salary from salaries s,employees as e where s.emp_no=e.emp_no and s.from_date=e.hire_date)as t2 where t1.emp_no=t2.emp_no order by growth; 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,d.dept_name, sum from departments as d, (select dept_no,sum(t)as sum from dept_emp as de, (select emp_no,count(*) as t from salaries group by emp_no)as e where e.emp_no = de.emp_no group by dept_no) as t1 where d.dept_no = t1.dept_no; 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; 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 d.dept_no, e.emp_no, s.salary from employees e, dept_emp d, salaries s where e.emp_no not in (select emp_no from dept_manager where to_date = '9999-01-01') and s.emp_no = e.emp_no and e.emp_no = d.emp_no and d.to_date = '9999-01-01' and s.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，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 t1.emp_no, t2.emp_no,t1.salary as s1,t2.salary as s2 from ( select salary,s.emp_no,dept_no from salaries as s join dept_emp de on s.emp_no=de.emp_no where s.to_date='9999-01-01' and de.to_date='9999-01-01' and s.emp_no not in (select emp_no from dept_manager dm) ) as t1, ( select salary,s.emp_no,dept_no from salaries as s join dept_manager dm on s.emp_no=dm.emp_no where s.to_date='9999-01-01' and dm.to_date='9999-01-01' ) as t2 where s1>s2 and t1.dept_no = t2.dept_no 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 d.dept_no, d.dept_name, t.title, count(t.title) as count from departments d, dept_emp de, titles t where d.dept_np = de.dept_no and de.emp_no = t.emp_no and de.to_date = '9999-01-01' and t.to_date = '9999-01-01' group by d.dept_no, t.title; 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 s.emp_no, s.from_date, s.salary-s1.salary AS salary_growth FROM salaries AS s, salaries AS s1 WHERE (strftime('%Y', s.to_date) - strftime('%Y', s1.to_date)) = 1 AND s.emp_no = s1.emp_no AND salary_growth > 5000 ORDER BY salary_growth DESC 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);相关操作
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(*) FROM category AS c, film_category AS f1, film AS f2 WHERE f1.film_id = f2.film_id AND c.category_id = f1.category_id AND f2.description LIKE '%robot%' GROUP BY f1.category_id HAVING COUNT(*) > 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 f left join film_category 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 title, description FROM film WHERE film_id in( SELECT film_id FROM film_category WHERE category_id in( SELECT category_id FROM category WHERE name='Action' ) )