目录
查找最晚入职员工的所有信息
查找入职员工时间排名倒数第三的员工所有信息
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
查找所有已经分配部门的员工的last_name和first_name
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
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
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
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
获取所有非manager的员工emp_no
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。
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
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。
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title进行忽略。
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
统计出当前各个title类型对应的员工当前(to_date='9999-01-01')薪水对应的平均工资。结果给出title以及平均工资avg。
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
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
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
查找员工编号emp_no为10001其自入职以来的薪水salary涨幅值growth
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
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
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
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'
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
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
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)
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
使用join查询方式找出没有分类的电影id以及名称
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
获取select * from employees对应的执行计划
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
创建一个actor表,包含如下列信息
对于表actor批量插入如下数据
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
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:
针对如下表actor结构创建索引:
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:
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。
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'
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。
删除emp_no重复的记录,只保留最小的id对应的记录。
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。
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。
将titles_test表名修改为titles_2017。
在audit表上创建外键约束,其emp_no对应employees_test表的主键id。
如何获取emp_v和employees有相同的数据?
将所有获取奖金的员工当前的薪水增加10%。
针对库中的所有表生成select count(*)对应的SQL语句
将employees表中的所有员工的last_name和first_name通过(')连接起来。
查找字符串'10,A,B' 中逗号','出现的次数cnt。
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
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
查找排除当前最大、最小salary之后的员工的平均工资avg_salary。
分页查询employees表,每5行一页,返回第2页的数据
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received ,没有分配具体的员工不显示
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。
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。
获取有奖金的员工相关信息。
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。。
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 * from employees
where hire_date =
(select max(hire_date) from employees)
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
where hire_date = (
select distinct hire_date from employees order by hire_date desc limit 2,1
)
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。
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`));
select s.*,d.dept_no
from salaries s,dept_manager d
where d.emp_no = s.emp_no and d.to_date = '9999-01-01' and s.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`));
select employees.last_name, first_name, dept_emp.dept_no
from dept_emp inner join employees
on dept_emp.emp_no = employees.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 `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 ep.last_name, ep.first_name, dp.dept_no
FROM employees ep
LEFT JOIN dept_emp dp
ON ep.emp_no = dp.emp_no
注意:
INNER JOIN 两边表同时有对应的数据,即任何一边缺失数据就不显示。
LEFT JOIN 会读取左边数据表的全部数据,即便右边表无对应数据。
RIGHT JOIN 会读取右边数据表的全部数据,即便左边表无对应数据。
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`));
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 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;
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 emp_no, COUNT(emp_no) AS t FROM salaries
GROUP BY emp_no HAVING t > 15
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 salary from salaries
where to_date='9999-01-01'
group by salary order by salary desc
大表一般用distinct效率不高,大数据量的时候都禁止用distinct,建议用group by解决重复问题。
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`));
方法一:使用NOT IN选出在employees但不在dept_manager中的emp_no记录
SELECT emp_no FROM employees
WHERE emp_no NOT IN (SELECT emp_no FROM dept_manager)
方法二:先使用LEFT JOIN连接两张表,再从此表中选出dept_no值为NULL对应的emp_no记录
SELECT emp_no FROM (SELECT * FROM employees LEFT JOIN dept_manager
ON employees.emp_no = dept_manager.emp_no)
WHERE dept_no IS NULL
方法三:方法二的简版,使用单层SELECT语句即可
SELECT employees.emp_no FROM employees LEFT JOIN dept_manager
ON employees.emp_no = dept_manager.emp_no
WHERE dept_no IS 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`));
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';
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 `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`));
此题思路如下:
1、先用INNER JOIN连接两张表,限制条件是两张表的emp_no相同,即d.emp_no = s.emp_no;
2、选取每个员工当前的工资水平,用d.to_date = '9999-01-01' AND s.to_date = '9999-01-01'作条件限制,因为此表中每条最新记录的 to_date 都用 9999-01-01 表示;
3、用GROUP BY d.dept_no将每个部门分为一组,用MAX()函数选取每组中工资最高者;
4、将salaries用s代替,dept_emp用d代替,最后将MAX(s.salary)用salary代替后输出。
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
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);
SELECT title,COUNT(title) AS t
FROM titles
GROUP BY title
HAVING t > 1;
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);
select title, count(distinct emp_no) as t
from titles
group by title
having t >= 2
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
where emp_no % 2 = 1
and last_name != 'Mary'
order by hire_date desc
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 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);
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
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 emp_no, salary from salaries
where to_date = '9999-01-01' and salary = (select distinct salary from salaries order by salary desc limit 1,1)
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 AS e INNER JOIN salaries AS s
ON e.emp_no = s.emp_no
WHERE s.to_date = '9999-01-01'
AND s.salary NOT IN (SELECT MAX(salary) FROM salaries WHERE to_date = '9999-01-01')
CREATE TABLE `departments` (
`dept_no` char(4) NOT NULL,
`dept_name` varchar(40) NOT NULL,
PRIMARY KEY (`dept_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 `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 em.last_name, em.first_name, dp.dept_name
FROM (employees AS em LEFT JOIN dept_emp AS de ON em.emp_no = de.emp_no)
LEFT JOIN departments AS dp ON de.dept_no = dp.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`));
SELECT (MAX(salary)-MIN(salary)) AS growth
FROM salaries WHERE emp_no = '10001'
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 a.emp_no, (b.salary - c.salary) as growth
from
employees as a
inner join salaries as b
on a.emp_no = b.emp_no and b.to_date = '9999-01-01'
inner join salaries as c
on a.emp_no = c.emp_no and a.hire_date = c.from_date
order by growth asc
CREATE TABLE `departments` (
`dept_no` char(4) NOT NULL,
`dept_name` varchar(40) NOT NULL,
PRIMARY KEY (`dept_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 `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.dept_no, a.dept_name, count(c.emp_no) as sum
from departments as a inner join dept_emp as b on a.dept_no = b.dept_no
inner join salaries as c on b.emp_no = c.emp_no
group by b.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`));
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;
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`));
SELECT de.dept_no, s.emp_no, s.salary
FROM dept_emp AS de INNER JOIN salaries AS s ON s.emp_no = de.emp_no AND s.to_date = '9999-01-01'
WHERE de.emp_no NOT IN (SELECT emp_no FROM dept_manager WHERE 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 `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`));
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
CREATE TABLE `departments` (
`dept_no` char(4) NOT NULL,
`dept_name` varchar(40) NOT NULL,
PRIMARY KEY (`dept_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 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);
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
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 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;
film表
字段 | 说明 |
film_id | 电影id |
title | 电影名称 |
description | 电影描述信息 |
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
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 |
CREATE TABLE category (
category_id tinyint(3) NOT NULL ,
name varchar(25) NOT NULL, `last_update` timestamp,
PRIMARY KEY ( category_id ));
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 |
CREATE TABLE film_category (
film_id smallint(5) NOT NULL,
category_id tinyint(3) NOT NULL, `last_update` timestamp);
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
film表
字段 | 说明 |
film_id | 电影id |
title | 电影名称 |
description | 电影描述信息 |
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
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 |
CREATE TABLE category (
category_id tinyint(3) NOT NULL ,
name varchar(25) NOT NULL, `last_update` timestamp,
PRIMARY KEY ( category_id ));
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 |
CREATE TABLE film_category (
film_id smallint(5) NOT NULL,
category_id tinyint(3) NOT NULL, `last_update` timestamp);
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
film表
字段 | 说明 |
film_id | 电影id |
title | 电影名称 |
description | 电影描述信息 |
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
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 |
CREATE TABLE category (
category_id tinyint(3) NOT NULL ,
name varchar(25) NOT NULL, `last_update` timestamp,
PRIMARY KEY ( category_id ));
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 |
CREATE TABLE film_category (
film_id smallint(5) NOT NULL,
category_id tinyint(3) NOT NULL, `last_update` timestamp);
select f.title,f.description from film as f
where f.film_id in (select fc.film_id from film_category as fc
where fc.category_id in (select c.category_id from category as c
where c.name = 'Action'));
explain模拟优化器执行SQL语句,在5.6以及以后的版本中,除过select,其他比如insert,update和delete均可以使用explain查看执行计划,从而知道mysql是如何处理sql语句,分析查询语句或者表结构的性能瓶颈。
作用
1、表的读取顺序
2、数据读取操作的操作类型
3、哪些索引可以使用
4、哪些索引被实际使用
5、表之间的引用
6、每张表有多少行被优化器查询
explain select * from employees
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 last_name||" "||first_name AS Name FROM employees
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)
)
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')
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');
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 |
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
select first_name,last_name from actor;
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);
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
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'
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')));
alter table actor
add `create_date` datetime not null default '0000-00-00 00:00:00'
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
);
构造触发器时注意以下几点:
1、用 CREATE TRIGGER 语句构造触发器,用 BEFORE或AFTER 来指定在执行后面的SQL语句之前或之后来触发TRIGGER
2、触发器执行的内容写出 BEGIN与END 之间
3、可以使用 NEW与OLD 关键字访问触发后或触发前的employees_test表单记录
create trigger audit_log after insert on employees_test
begin
insert into audit values(new.id,new.name);
end;
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)
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');
set 字段名1=字段值1,字段名2=字段值2 ,用逗号","分隔,不是用空格分隔
update titles_test set to_date=null ,from_date ='2001-01-01'
where to_date='9999-01-01'
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');
replace into titles_test values(5, 10005, 'Senior Engineer', '1986-06-26', '9999-01-01');
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');
alter table titles_test rename to titles_2017
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
);
alter table audit
add foreign key(emp_no) references employees_test(id)
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`));
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;
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)
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'
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
select (length('10,A,B')-length(replace('10,A,B',',',''))) as cnt
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)
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;
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 |
---|
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);
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
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`));
create table emp_bonus(
emp_no int not null,
recevied datetime not null,
btype smallint not null);
输出格式:
e.emp_no | dept_no | btype | received |
---|---|---|---|
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
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);
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`));
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;
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';
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 |
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
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;