sql-练习

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  1. SQL 练习

SQL 练习

# 员工表
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`));
  1. 查找最晚入职的员工的所有信息
  • 日期越大，越晚入职，max函数
select * from employees where hire_date = (select max(hire_date) from employees)
  1. 查找入职员工时间排名倒数第三的员工所有信息
  • order by, distinct, desc limit 的使用
select * from employees where hire_date = (
    select distinct hire_date from employees order by hire_date desc limit 2,1
)

# 部门经理表
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`));
# 工资表
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`));
  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
  • 联表查询
select s.* ,d.dept_no from salaries as s join dept_manager  as d 
    on s.emp_no = d.emp_no 
    where s.to_date = '9999-01-01' and d.to_date = '9999-01-01'

# 部门员工表
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`));
# 职工表
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`));
  1. 查找所有已经分配部门的员工的last_name和first_name
  • inner join … on …
select e.last_name,e.first_name,de.dept_no
    from dept_emp as de
    inner join employees as e 
    on de.emp_no = e.emp_no;
  1. 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
  • 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。
select e.last_name,e.first_name,de.dept_no
    from employees as e 
    left join dept_emp as de 
    on de.emp_no = e.emp_no;
  1. 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
# way 1
select e.emp_no,s.salary 
    from employees as e
    inner join salaries as s
    on e.hire_date = s.from_date and e.emp_no = s.emp_no
    order by e.emp_no desc;

# way 2
select e.emp_no,s.salary
    from employees as e,salaries as s
    where e.emp_no = s.emp_no and e.hire_date = s.from_date
    order by e.emp_no desc
  1. 查找薪水变化超过15次的员工号emp_no以及其对应的变化次数t
  • group by 属性 having 条件
select emp_no,count(emp_no) as t from salaries
    group by emp_no having t > 15;
  1. 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
  • 对于distinct与group by的使用:
    • 当对系统的性能高并数据量大时使用group by
    • 当对系统的性能不高时使用数据量少时两者皆可
    • 尽量使用group by
select distinct salary from salaries
    where to_date = '9999-01-01'
    order by salary desc

select salary from salaries
    where to_date = '9999-01-01'
    group by salary
    order by salary desc
  1. 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 dept_manager as d
    inner join salaries as s
    on d.emp_no = s.emp_no
    and d.to_date = '9999-01-01' 
    and s.to_date = '9999-01-01'
    group by d.emp_no;
  1. 获取所有非manager的员工emp_no
  • not in
# use : not in
select e.emp_no 
    from employees as e
    where e.emp_no not in (
    select d.emp_no from dept_manager as d
)
# 使用 left join 和 NULL
SELECT employees.emp_no FROM employees 
    LEFT JOIN dept_manager
    ON employees.emp_no = dept_manager.emp_no
WHERE dept_no IS NULL
  1. 获取员工当前的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 AS manager_no 
    from dept_emp as de inner join dept_manager as dm 
    on de.dept_no = dm.dept_no -- 部门一致
    -- 时间期限限定 manager自己不用显示（员工表的员工号不应该与部门表的员工号一致）
    where dm.to_date = '9999-01-01' AND de.to_date = '9999-01-01' AND de.emp_no <> dm.emp_no
  1. 获取所有部门中当前员工薪水最高的相关信息
  • 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,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 的分组功能，分了组就会取每组的最大值
    group by d.dept_no
  1. 从titles表获取按照title进行分组
    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);

select title,count(title) as t 
    from titles
    group by title
    -- 条件：title的数量大于等于2
    -- 由于WHERE后不可跟COUNT()函数，故用HAVING语句来限定t>=2的条件
    having t >= 2
  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。
  • 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。
-- distinct emp_no 避免emp_no重复
select title,count(distinct emp_no) as t 
    from titles
    group by title
    -- 条件：title的数量大于等于2
    -- 由于WHERE后不可跟COUNT()函数，故用HAVING语句来限定t>=2的条件
    having t >= 2
  1. 查找employees表
  • 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
    -- 字符串 要用: ''
    where emp_no % 2 = 1 and last_name != 'Mary'
    -- order by XXX desc 表示按照 XXX 逆序
    order by hire_date desc
  1. 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
  • 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。
select t.title,avg(s.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
  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
  • 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 * from table limit [m],n;
  • 其中
    • m—— [m]为可选，如果填写表示skip步长，即跳过m条。
    • n——显示条数。指从第m+1条记录开始，取n条记录。
select emp_no,salary
    from salaries
    where to_date  = '9999-01-01' and 
    -- order by 排序
    -- desc 逆序
    -- limit 限制从第几位开始
    salary = (select distinct salary from salaries order by salary desc limit 1,1)
  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，不准使用order by
  • 多级查询，
  • 不等于号 <>
  • max函数
select e.emp_no,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 to_date = '9999-01-01' and 
        salary = (
        select max(salary) from salaries
        where salary <> (select max(salary) from salaries where to_date = '9999-01-01')
        and to_date = '9999-01-01'
    )
  1. 查找所有员工的last_name和first_name以及对应的dept_name
  • 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,dp.dept_name 
    from (employees as e left join dept_emp as de on e.emp_no = de.emp_no)
    left join departments as dp on dp.dept_no = de.dept_no
  1. 查找员工编号emp_no为10001其自入职以来的薪水salary涨幅值growth
  • min 和 max
-- way 1
select (max(salary) - min(salary)) as growth 
    from salaries where emp_no = '10001'

-- way 2
select (    
    (select salary from salaries where emp_no = '10001' order by to_date  desc limit 1) -
    (select salary from salaries where emp_no = '10001' order by to_date asc limit 1)
    ) 
    as growth
  1. 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
  • 综合使用left join
  • order by
  • 根据所给表格，找出查找所需数据的条件
select sCurrent.emp_no,(sCurrent.salary-sStart.salary) as growth
from (select s.emp_no,s.salary from employees e left join salaries 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 e left join salaries 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
  1. 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
  • 按照部门id分组，查询所有员工的工资
-- 简单解法
select dm.dept_no,dm.dept_name,count(*) as sum
    from departments as dm,dept_emp as de,salaries as s
    where s.emp_no = de.emp_no and dm.dept_no = de.dept_no
    group by dm.dept_no
-- 利用join，和上面一样的？
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
  1. 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
  • 按照薪水排名
  • 按照emp_no排名
  • 排名累计
-- 求排名： s1.salary <= s2.salary，意思是在输出s1.salary的情况下，有多少个s2.salary大于等于s1.salary，
select s1.emp_no,s1.salary,count(distinct s2.salary) as rank
from salaries as s1,salaries as s2
-- 找到满足时间条件的数据，和排序数据条件
where s1.to_date = '9999-01-01' and s2.to_date = '9999-01-01' and s1.salary <= s2.salary
-- 按照emp_no 进行分组
group by s1.emp_no
-- 进行排序，按照两个属性排序
order by s1.salary desc,s1.emp_no asc
  1. 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’
  • 一步一步来
    • 先找出所有满足时间条件的工资数据
    • 然后将工资数据与部门no绑定
    • 去除其中的manager的工资数据
select de.dept_no,s.emp_no,s.salary
from (employees as e inner join salaries as s on s.emp_no = e.emp_no and s.to_date = '9999-01-01')
inner join dept_emp as de on e.emp_no = de.emp_no
where de.emp_no not in (select emp_no from dept_manager where to_date = '9999-01-01')
  1. 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

  • 先获取所有员工的工资、和部门信息

  • 再获取所有经理的工资、和部门信息

  • 将部门相同的员工和经理的工资进行比较，求员工工资高于经理工资的信息

select des.emp_no as emp_no,dms.emp_no as manager_no,des.salary as emp_salary,dms.salary as manager_salary 
from
    (select s.emp_no,s.salary,de.dept_no 
     from salaries as s 
     inner join dept_emp as de 
     on s.emp_no = de.emp_no and s.to_date = '9999-01-01') as des,
    (select s.emp_no,s.salary,dm.dept_no 
     from salaries as s 
     inner join dept_manager as dm 
     on s.emp_no = dm.emp_no and s.to_date = '9999-01-01') as dms
where des.dept_no = dms.dept_no and des.salary > dms.salary
  1. 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
  • 按照部门和title来统计数量
select de.dept_no,d.dept_name,t.title,count(t.title) as count 
from departments as d,dept_emp as de,titles  as t
where d.dept_no = 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
  • 使用inner join
select de.dept_no,dp.dapt_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
  1. 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 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
-- 工资涨幅
and salary_growth > 5000
-- 时间限制
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
  1. 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
  • 包括robot
  • 电影数量得不低于5部
  • 电影名称、电影分类、电影数目
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);
  • 模糊查询
  • 先找到出现次数不低于5的分类
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,file_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
  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 right join film_category as fc
-- 因为结果还没出来，所以fc.category_id is null 无效
on fc.category_id is null and f.film_id = fc.film_id

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
  1. 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
-- 找到满足分类的分类id
select category_id from category where name = 'Action' as cid

select f.title,f.description from (
    -- 找到film_id
    select film_id from (
        select category_id from category where name = 'Action'
    ) as cid, film_category as fc where fc.category_id = cid.category_id
) as fd,film as f
where f.film_id = fd.film_id
转载请注明来源，欢迎对文章中的引用来源进行考证，欢迎指出任何有错误或不够清晰的表达。可以在下面评论区评论，也可以邮件至 1056615746@qq.com

💰

Title:sql-练习

Count:3.4k

Author:攀登

Created At:2019-12-27, 23:30:40

Updated At:2024-06-15, 15:52:32

Url:http://jiafeimao-gjf.github.io/2019/12/27/sql-practise/

Copyright: 'Attribution-non-commercial-shared in the same way 4.0' Reprint please keep the original link and author.

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