目录
1. 175组合两个表
左外连接
2. 181. 超过经理收入的员工
3. 182. 查找重复的电子邮箱
4. 196. 删除重复的电子邮箱
5. 197. 上升的温度
日期作差
6. 511. 游戏玩法分析 I
7. 577. 员工奖金
null条件运算
8. 584. 寻找用户推荐人
9. 586. 订单最多的客户
10. 595. 大的国家
11. 596. 超过5名学生的课
聚合加条件
12. 607. 销售员
13. 610. 判断三角形
if语句
14. 619. 只出现一次的最大数字
仅独立使用select
15. 620. 有趣的电影
16. 627.update替换
17. 1050. 合作过至少三次的演员和导演
group by两个参数
18. 1068. 产品销售分析 I
19. 1084. 销售分析III
日期比较
20. 1075. 项目员工 I
小数点位数 round(x,2)
21. 1141. 查询近30天活跃用户数
22. 1148. 文章浏览 I
23. 1179. 重新格式化部门表
行转列暴力解法
24. 1211. 查询结果的质量和占比
分组后求算式
25. 1251. 平均售价
ifnull(算式,空时的值)
26. 1280. 学生们参加各科测试的次数
连接多
27. 1327. 列出指定时间段内所有的下单产品
日期函数YEAR()
28. 1378. 使用唯一标识码替换员工ID
29. 1407. 排名靠前的旅行者
陷阱
30. 1484. 按日期分组销售产品
group_concat(distinct product)
31. 1527. 患某种疾病的患者
字符串占位符
32. 1587. 银行账户概要 II
33. 1581. 进店却未进行过交易的顾客
null统计不到个数,要使用count(*)
34. 1517. 查找拥有有效邮箱的用户
正则表达式!!
35. 1633. 各赛事的用户注册率
36. 1661. 每台机器的进程平均运行时间
37. 1667. 修复表中的名字
38. 1683. 无效的推文
39. 1693. 每天的领导和合伙人
40. 1729. 求关注者的数量
41. 1731. 每位经理的下属员工数量
42. 1741. 查找每个员工花费的总时间
43. 1757. 可回收且低脂的产品
44. 1789. 员工的直属部门
union
45. 1795. 每个产品在不同商店的价格
46. 1873. 计算特殊奖金
if嵌套
47. 1890. 2020年最后一次登录
48.1965. 丢失信息的雇员
49. 1978. 上级经理已离职的公司员工
50. 2356. 每位教师所教授的科目种类的数量
Q:
表: Person
+-------------+---------+
| 列名 | 类型 |
+-------------+---------+
| PersonId | int |
| FirstName | varchar |
| LastName | varchar |
+-------------+---------+
personId 是该表的主键(具有唯一值的列)。
该表包含一些人的 ID 和他们的姓和名的信息。
表: Address
+-------------+---------+
| 列名 | 类型 |
+-------------+---------+
| AddressId | int |
| PersonId | int |
| City | varchar |
| State | varchar |
+-------------+---------+
addressId 是该表的主键(具有唯一值的列)。
该表的每一行都包含一个 ID = PersonId 的人的城市和州的信息。
编写解决方案,报告 Person 表中每个人的姓、名、城市和州。如果 personId 的地址不在 Address 表中,则报告为 null 。
以 任意顺序 返回结果表。
结果格式如下所示。
示例 1:
输入:
Person表:
+----------+----------+-----------+
| personId | lastName | firstName |
+----------+----------+-----------+
| 1 | Wang | Allen |
| 2 | Alice | Bob |
+----------+----------+-----------+
Address表:
+-----------+----------+---------------+------------+
| addressId | personId | city | state |
+-----------+----------+---------------+------------+
| 1 | 2 | New York City | New York |
| 2 | 3 | Leetcode | California |
+-----------+----------+---------------+------------+
输出:
+-----------+----------+---------------+----------+
| firstName | lastName | city | state |
+-----------+----------+---------------+----------+
| Allen | Wang | Null | Null |
| Bob | Alice | New York City | New York |
+-----------+----------+---------------+----------+
解释:
地址表中没有 personId = 1 的地址,所以它们的城市和州返回 null。
addressId = 1 包含了 personId = 2 的地址信息。
SELECT firstName, lastName, city, state from Person left join Address on Person.personId=Address.personId;
2. 181. 超过经理收入的员工
表:Employee
+-------------+---------+
| Column Name | Type |
+-------------+---------+
| id | int |
| name | varchar |
| salary | int |
| managerId | int |
+-------------+---------+
id 是该表的主键(具有唯一值的列)。
该表的每一行都表示雇员的ID、姓名、工资和经理的ID。
编写解决方案,找出收入比经理高的员工。
以 任意顺序 返回结果表。
结果格式如下所示。
示例 1:
输入:
Employee 表:
+----+-------+--------+-----------+
| id | name | salary | managerId |
+----+-------+--------+-----------+
| 1 | Joe | 70000 | 3 |
| 2 | Henry | 80000 | 4 |
| 3 | Sam | 60000 | Null |
| 4 | Max | 90000 | Null |
+----+-------+--------+-----------+
输出:
+----------+
| Employee |
+----------+
| Joe |
+----------+
解释: Joe 是唯一挣得比经理多的雇员。
SELECT a.name Employee FROM Employee a join Employee b on a.managerId=b.id where a.salary>b.salary;
输入:
Person 表:
+----+---------+
| id | email |
+----+---------+
| 1 | [email protected] |
| 2 | [email protected] |
| 3 | [email protected] |
+----+---------+
输出:
+---------+
| Email |
+---------+
| [email protected] |
+---------+
解释: [email protected] 出现了两次。
select email as Email from Person group by email having count(email) > 1
4. 183. 从不订购的客户
select t.name as Customers from (select C.id as i,name,O.id as ii,customerId from Customers as C left join Orders as O on C.id=O.customerId) as t where t.customerId is null
or
select name as Customers from Customers as C left join Orders as O on C.id=O.customerId where O.customerId is null
编写解决方案 删除 所有重复的电子邮件,只保留一个具有最小 id
的唯一电子邮件。
(对于 SQL 用户,请注意你应该编写一个 DELETE
语句而不是 SELECT
语句。)
| id | email |
| -- | ---------------- |
| 1 | [email protected] |
| 2 | [email protected] |
| 3 | [email protected] |
DELETE FROM Person
WHERE id NOT IN (
# 筛出所有的单个最小id,删掉不在里面的
SELECT id FROM (
SELECT MIN(id) AS id FROM Person GROUP BY email
) AS u
);
编写解决方案,找出与之前(昨天的)日期相比温度更高的所有日期的 id
。
返回结果 无顺序要求 。
结果格式如下例子所示。
# Write your MySQL query statement below
select t.id1 id from (
select w1.id id1,w1.recordDate re1,w1.temperature te1,w2.id id2,w2.recordDate re2,w2.temperature te2
from Weather as w1 left join Weather as w2 on TIMESTAMPDIFF(DAY,w2.recordDate, w1.recordDate)=1
) as t where t.te1>t.te2
查询每位玩家 第一次登陆平台的日期。
查询结果的格式如下所示:
Activity 表:
+-----------+-----------+------------+--------------+
| player_id | device_id | event_date | games_played |
+-----------+-----------+------------+--------------+
| 1 | 2 | 2016-03-01 | 5 |
| 1 | 2 | 2016-05-02 | 6 |
| 2 | 3 | 2017-06-25 | 1 |
| 3 | 1 | 2016-03-02 | 0 |
| 3 | 4 | 2018-07-03 | 5 |
+-----------+-----------+------------+--------------+
Result 表:
+-----------+-------------+
| player_id | first_login |
+-----------+-------------+
| 1 | 2016-03-01 |
| 2 | 2017-06-25 |
| 3 | 2016-03-02 |
+-----------+-------------+
# Write your MySQL query statement below
select player_id,min(event_date) first_login from(
select player_id,event_date from Activity
) as t group by player_id
编写解决方案,报告每个奖金 少于 1000
的员工的姓名和奖金数额。
结果格式如下所示。
示例 1:
输入:
Employee table:
+-------+--------+------------+--------+
| empId | name | supervisor | salary |
+-------+--------+------------+--------+
| 3 | Brad | null | 4000 |
| 1 | John | 3 | 1000 |
| 2 | Dan | 3 | 2000 |
| 4 | Thomas | 3 | 4000 |
+-------+--------+------------+--------+
Bonus table:
+-------+-------+
| empId | bonus |
+-------+-------+
| 2 | 500 |
| 4 | 2000 |
+-------+-------+
输出:
+------+-------+
| name | bonus |
+------+-------+
| Brad | null |
| John | null |
| Dan | 500 |
+------+-------+
select name,bonus from Employee as e left join Bonus as b
on e.empId=b.empId
where b.bonus < 1000 or b.bonus is null
Q:在这个语句里我筛选出了bonus小于1000的数据,但是bonus为null为什么筛选不出来?
A: 这是因为在MySQL中,当你使用条件筛选时,NULL 值的处理方式会有些不同。通常情况下,比较 NULL 值需要使用
IS NULL
或IS NOT NULL
,而不是使用比较运算符(如<
、>
、=
)。
找出那些 没有被 id = 2 的客户 推荐 的客户的姓名。
以 任意顺序 返回结果表。
结果格式如下所示。
示例 1:
输入:
Customer 表:
+----+------+------------+
| id | name | referee_id |
+----+------+------------+
| 1 | Will | null |
| 2 | Jane | null |
| 3 | Alex | 2 |
| 4 | Bill | null |
| 5 | Zack | 1 |
| 6 | Mark | 2 |
+----+------+------------+
easy
查找下了 最多订单 的客户的 customer_number 。
测试用例生成后, 恰好有一个客户 比任何其他客户下了更多的订单。
查询结果格式如下所示。
示例 1:
输入:
Orders 表:
+--------------+-----------------+
| order_number | customer_number |
+--------------+-----------------+
| 1 | 1 |
| 2 | 2 |
| 3 | 3 |
| 4 | 3 |
+--------------+-----------------+
输出:
+-----------------+
| customer_number |
+-----------------+
| 3 |
+-----------------+
select customer_number
from orders
group by customer_number
order by count(*) desc
limit 1
如果一个国家满足下述两个条件之一,则认为该国是 大国 :
3000000 km2
),或者25000000
)送分题查询 至少有5个学生 的所有班级。
以 任意顺序 返回结果表。
查询结果格式如下所示。
示例 1:
输入:
Courses table:
+---------+----------+
| student | class |
+---------+----------+
| A | Math |
| B | English |
| C | Math |
| D | Biology |
| E | Math |
| F | Computer |
| G | Math |
| H | Math |
| I | Math |
+---------+----------+
输出:
+---------+
| class |
+---------+
| Math |
+---------+
select class from Courses
group by class
having count(student)>=5
编写解决方案,找出没有任何与名为 “RED” 的公司相关的订单的所有销售人员的姓名。
以 任意顺序 返回结果表。
返回结果格式如下所示。
示例 1:
输入:
SalesPerson 表:
+----------+------+--------+-----------------+------------+
| sales_id | name | salary | commission_rate | hire_date |
+----------+------+--------+-----------------+------------+
| 1 | John | 100000 | 6 | 4/1/2006 |
| 2 | Amy | 12000 | 5 | 5/1/2010 |
| 3 | Mark | 65000 | 12 | 12/25/2008 |
| 4 | Pam | 25000 | 25 | 1/1/2005 |
| 5 | Alex | 5000 | 10 | 2/3/2007 |
+----------+------+--------+-----------------+------------+
Company 表:
+--------+--------+----------+
| com_id | name | city |
+--------+--------+----------+
| 1 | RED | Boston |
| 2 | ORANGE | New York |
| 3 | YELLOW | Boston |
| 4 | GREEN | Austin |
+--------+--------+----------+
Orders 表:
+----------+------------+--------+----------+--------+
| order_id | order_date | com_id | sales_id | amount |
+----------+------------+--------+----------+--------+
| 1 | 1/1/2014 | 3 | 4 | 10000 |
| 2 | 2/1/2014 | 4 | 5 | 5000 |
| 3 | 3/1/2014 | 1 | 1 | 50000 |
| 4 | 4/1/2014 | 1 | 4 | 25000 |
+----------+------------+--------+----------+--------+
输出:
+------+
| name |
+------+
| Amy |
| Mark |
| Alex |
+------+
select tem.name name from (
select s.sales_id id,s.name name,o.com_id com_id from SalesPerson as s left join Orders as o on s.sales_id=o.sales_id
) as tem left join Company as c on tem.com_id=c.com_id or tem.name is null
group by tem.id
having sum(if(c.name='RED',1,0))=0
输入:
Triangle 表:
+----+----+----+
| x | y | z |
+----+----+----+
| 13 | 15 | 30 |
| 10 | 20 | 15 |
+----+----+----+
输出:
+----+----+----+----------+
| x | y | z | triangle |
+----+----+----+----------+
| 13 | 15 | 30 | No |
| 10 | 20 | 15 | Yes |
+----+----+----+----------+
select x,y,z,(if(x+y>z and x+z>y and y+z>x,'Yes','No')) as triangle from Triangle;
单一数字 是在 MyNumbers 表中只出现一次的数字。
找出最大的 单一数字 。如果不存在 单一数字 ,则返回 null 。
查询结果如下例所示。
示例 1:
输入:
MyNumbers 表:
+-----+
| num |
+-----+
| 8 |
| 8 |
| 3 |
| 3 |
| 1 |
| 4 |
| 5 |
| 6 |
+-----+
输出:
+-----+
| num |
+-----+
| 6 |
+-----+
还有null的情况
# Write your MySQL query statement below
select (
select num from MyNumbers
group by num
having count(num)=1
order by num desc
limit 1
) as num
select null as num
| num |
| ---- |
| null |
编写解决方案,找出所有影片描述为 非 boring
(不无聊) 的并且 id 为奇数 的影片。
返回结果按 rating
降序排列。
结果格式如下示例。送分题
select * from cinema where id%2=1 and description!='boring'
order by rating desc
请你编写一个解决方案来交换所有的 'f'
和 'm'
(即,将所有 'f'
变为 'm'
,反之亦然),仅使用 单个 update 语句 ,且不产生中间临时表。
注意,你必须仅使用一条 update 语句,且 不能 使用 select 语句。
update Salary
set sex = if(sex='m','f','m')
编写解决方案找出合作过至少三次的演员和导演的 id 对 (actor_id, director_id)
# Write your MySQL query statement below
select Ac.actor_id,Ac.director_id from ActorDirector as Ac
group by Ac.actor_id,Ac.director_id
having count(*)>=3
送分题
select p.product_name,s.year,s.price from Sales as s left join Product as p on s.product_id=p.product_id
即仅在2019-01-01
至2019-03-31
(含)之间出售的商品。
# Write your MySQL query statement below
select t.product_id,p.product_name from (
select s.product_id from Sales as s
group by s.product_id having max(s.sale_date)<'2019-04-01' and min(s.sale_date)>='2019-01-01'
) as t left join Product as p on t.product_id=p.product_id
Project 表:
+-------------+-------------+
| project_id | employee_id |
+-------------+-------------+
| 1 | 1 |
| 1 | 2 |
| 1 | 3 |
| 2 | 1 |
| 2 | 4 |
+-------------+-------------+
Employee 表:
+-------------+--------+------------------+
| employee_id | name | experience_years |
+-------------+--------+------------------+
| 1 | Khaled | 3 |
| 2 | Ali | 2 |
| 3 | John | 1 |
| 4 | Doe | 2 |
+-------------+--------+------------------+
Result 表:
+-------------+---------------+
| project_id | average_years |
+-------------+---------------+
| 1 | 2.00 |
| 2 | 2.50 |
+-------------+---------------+
第一个项目中,员工的平均工作年限是 (3 + 2 + 1) / 3 = 2.00;第二个项目中,员工的平均工作年限是 (3 + 2) / 2 = 2.50
select p.project_id,round(avg(e.experience_years),2) as average_years from Project as p join Employee as e on p.employee_id=e.employee_id
group by p.project_id
编写解决方案,统计截至 2019-07-27
(包含2019-07-27),近 30
天的每日活跃用户数(当天只要有一条活动记录,即为活跃用户)
输入:
Activity table:
+---------+------------+---------------+---------------+
| user_id | session_id | activity_date | activity_type |
+---------+------------+---------------+---------------+
| 1 | 1 | 2019-07-20 | open_session |
| 1 | 1 | 2019-07-20 | scroll_down |
| 1 | 1 | 2019-07-20 | end_session |
| 2 | 4 | 2019-07-20 | open_session |
| 2 | 4 | 2019-07-21 | send_message |
| 2 | 4 | 2019-07-21 | end_session |
| 3 | 2 | 2019-07-21 | open_session |
| 3 | 2 | 2019-07-21 | send_message |
| 3 | 2 | 2019-07-21 | end_session |
| 4 | 3 | 2019-06-25 | open_session |
| 4 | 3 | 2019-06-25 | end_session |
+---------+------------+---------------+---------------+
输出:
+------------+--------------+
| day | active_users |
+------------+--------------+
| 2019-07-20 | 2 |
| 2019-07-21 | 2 |
+------------+--------------+
select day,count(active_users) as active_users from (
select distinct a.activity_date as day,a.user_id as active_users from Activity as a
where a.activity_date>'2019-06-27' and a.activity_date<'2019-07-28'
)as t
group by day
请查询出所有浏览过自己文章的作者
秒了
输入:
Views 表:
+------------+-----------+-----------+------------+
| article_id | author_id | viewer_id | view_date |
+------------+-----------+-----------+------------+
| 1 | 3 | 5 | 2019-08-01 |
| 1 | 3 | 6 | 2019-08-02 |
| 2 | 7 | 7 | 2019-08-01 |
| 2 | 7 | 6 | 2019-08-02 |
| 4 | 7 | 1 | 2019-07-22 |
| 3 | 4 | 4 | 2019-07-21 |
| 3 | 4 | 4 | 2019-07-21 |
+------------+-----------+-----------+------------+
输出:
+------+
| id |
+------+
| 4 |
| 7 |
+------+
select distinct v.viewer_id as id from Views as v
where v.author_id=v.viewer_id
order by id
SELECT
DISTINCT
a.id,
Jan.revenue AS Jan_Revenue,
Feb.revenue AS Feb_Revenue,
Mar.revenue AS Mar_Revenue,
Apr.revenue AS Apr_Revenue,
May.revenue AS May_Revenue,
Jun.revenue AS Jun_Revenue,
Jul.revenue AS Jul_Revenue,
Aug.revenue AS Aug_Revenue,
Sep.revenue AS Sep_Revenue,
Octo.revenue AS Oct_Revenue,
Nov.revenue AS Nov_Revenue,
Dece.revenue AS Dec_Revenue
FROM
Department a LEFT JOIN Department Jan ON a.id = Jan.id
AND
Jan.month = 'Jan'LEFT JOIN Department Feb ON a.id = Feb.id
AND
Feb.month = 'Feb' LEFT JOIN Department Mar ON a.id = Mar.id
AND
Mar.month = 'Mar' LEFT JOIN Department Apr ON a.id = Apr.id
AND
Apr.month = 'Apr' LEFT JOIN Department May ON a.id = May.id
AND
May.month = 'May' LEFT JOIN Department Jun ON a.id = Jun.id
AND
Jun.month = 'Jun' LEFT JOIN Department Jul ON a.id = Jul.id
AND
Jul.month = 'Jul' LEFT JOIN Department Aug ON a.id = Aug.id
AND
Aug.month = 'Aug' LEFT JOIN Department Sep ON a.id = Sep.id
AND
Sep.month = 'Sep' LEFT JOIN Department Octo ON a.id = Octo.id
AND
Octo.month = 'Oct'LEFT JOIN Department Nov ON a.id = Nov.id
AND
Nov.month = 'Nov' LEFT JOIN Department Dece ON a.id = Dece.id
AND
Dece.month = 'Dec'
当一个单元格中有多个数据时,case when只会提取当中的第一个数据。
那该如何解决单元格内含多个数据的情况呢?答案就是使用聚合函数,聚合函数就用来输入多个数据,输出一个数据的。如SUM()或MAX(),而每个聚合函数的输入就是每一个多数据的单元格。
将查询结果的质量 quality 定义为:
各查询结果的评分与其位置之间比率的平均值。
将劣质查询百分比 poor_query_percentage 为:
评分小于 3 的查询结果占全部查询结果的百分比。
输入:
Queries table:
+------------+-------------------+----------+--------+
| query_name | result | position | rating |
+------------+-------------------+----------+--------+
| Dog | Golden Retriever | 1 | 5 |
| Dog | German Shepherd | 2 | 5 |
| Dog | Mule | 200 | 1 |
| Cat | Shirazi | 5 | 2 |
| Cat | Siamese | 3 | 3 |
| Cat | Sphynx | 7 | 4 |
+------------+-------------------+----------+--------+
输出:
+------------+---------+-----------------------+
| query_name | quality | poor_query_percentage |
+------------+---------+-----------------------+
| Dog | 2.50 | 33.33 |
| Cat | 0.66 | 33.33 |
+------------+---------+-----------------------+
解释:
Dog 查询结果的质量为 ((5 / 1) + (5 / 2) + (1 / 200)) / 3 = 2.50
Dog 查询结果的劣质查询百分比为 (1 / 3) * 100 = 33.33
Cat 查询结果的质量为 ((2 / 5) + (3 / 3) + (4 / 7)) / 3 = 0.66
Cat 查询结果的劣质查询百分比为 (1 / 3) * 100 = 33.33
select query_name,round(avg(rating/position),2) as quality,round(sum(if(rating<3,1,0))/count(rating)*100,2) as poor_query_percentage
from Queries
group by query_name
计算题
示例 1:
输入:
Prices table:
+------------+------------+------------+--------+
| product_id | start_date | end_date | price |
+------------+------------+------------+--------+
| 1 | 2019-02-17 | 2019-02-28 | 5 |
| 1 | 2019-03-01 | 2019-03-22 | 20 |
| 2 | 2019-02-01 | 2019-02-20 | 15 |
| 2 | 2019-02-21 | 2019-03-31 | 30 |
+------------+------------+------------+--------+
UnitsSold table:
+------------+---------------+-------+
| product_id | purchase_date | units |
+------------+---------------+-------+
| 1 | 2019-02-25 | 100 |
| 1 | 2019-03-01 | 15 |
| 2 | 2019-02-10 | 200 |
| 2 | 2019-03-22 | 30 |
+------------+---------------+-------+
输出:
+------------+---------------+
| product_id | average_price |
+------------+---------------+
| 1 | 6.96 |
| 2 | 16.96 |
+------------+---------------+
解释:
平均售价 = 产品总价 / 销售的产品数量。
产品 1 的平均售价 = ((100 * 5)+(15 * 20) )/ 115 = 6.96
产品 2 的平均售价 = ((200 * 15)+(30 * 30) )/ 230 = 16.96
# Write your MySQL query statement below
select p.product_id as product_id,ifnull(round(sum(u.units*p.price) / sum(u.units),2),0) as average_price
from Prices as p left join UnitsSold as u
on u.purchase_date <= p.end_date and u.purchase_date>=p.start_date and u.product_id=p.product_id
group by p.product_id
查询出每个学生参加每一门科目测试的次数,结果按 student_id 和 subject_name 排序。
查询结构格式如下所示。
示例 1:
输入:
Students table:
+------------+--------------+
| student_id | student_name |
+------------+--------------+
| 1 | Alice |
| 2 | Bob |
| 13 | John |
| 6 | Alex |
+------------+--------------+
Subjects table:
+--------------+
| subject_name |
+--------------+
| Math |
| Physics |
| Programming |
+--------------+
Examinations table:
+------------+--------------+
| student_id | subject_name |
+------------+--------------+
| 1 | Math |
| 1 | Physics |
| 1 | Programming |
| 2 | Programming |
| 1 | Physics |
| 1 | Math |
| 13 | Math |
| 13 | Programming |
| 13 | Physics |
| 2 | Math |
| 1 | Math |
+------------+--------------+
输出:
+------------+--------------+--------------+----------------+
| student_id | student_name | subject_name | attended_exams |
+------------+--------------+--------------+----------------+
| 1 | Alice | Math | 3 |
| 1 | Alice | Physics | 2 |
| 1 | Alice | Programming | 1 |
| 2 | Bob | Math | 1 |
| 2 | Bob | Physics | 0 |
| 2 | Bob | Programming | 1 |
| 6 | Alex | Math | 0 |
| 6 | Alex | Physics | 0 |
| 6 | Alex | Programming | 0 |
| 13 | John | Math | 1 |
| 13 | John | Physics | 1 |
| 13 | John | Programming | 1 |
+------------+--------------+--------------+----------------+
解释:
结果表需包含所有学生和所有科目(即便测试次数为0):
Alice 参加了 3 次数学测试, 2 次物理测试,以及 1 次编程测试;
Bob 参加了 1 次数学测试, 1 次编程测试,没有参加物理测试;
Alex 啥测试都没参加;
John 参加了数学、物理、编程测试各 1 次。
select tem.student_id,tem.student_name,tem.subject_name,sum(if (tem.subject_name=cishu.subject_name,1,0)) as attended_exams from (
select stu.student_id as student_id,stu.student_name as student_name,sub.subject_name as subject_name from Students as stu join Subjects as sub
) as tem left join Examinations as cishu on tem.student_id=cishu.student_id and tem.subject_name=cishu.subject_name
group by tem.student_id,tem.subject_name
order by tem.student_id,tem.subject_name
select p.product_name,t.unit from (
select product_id,sum(unit) unit from Orders
where MONTH(order_date)=2 and YEAR(order_date)=2020
group by product_id
having sum(unit)>=100
)as t left join Products as p on t.product_id=p.product_id
秒了送分题
select uni.unique_id as unique_id,e.name from Employees as e left join EmployeeUNI as uni on e.id=uni.id
编写解决方案,报告每个用户的旅行距离。
返回的结果表单,以 travelled_distance 降序排列 ,如果有两个或者更多的用户旅行了相同的距离, 那么再以 name 升序排列 。
返回结果格式如下例所示。
示例 1:
输入:
Users 表:
+------+-----------+
| id | name |
+------+-----------+
| 1 | Alice |
| 2 | Bob |
| 3 | Alex |
| 4 | Donald |
| 7 | Lee |
| 13 | Jonathan |
| 19 | Elvis |
+------+-----------+
Rides 表:
+------+----------+----------+
| id | user_id | distance |
+------+----------+----------+
| 1 | 1 | 120 |
| 2 | 2 | 317 |
| 3 | 3 | 222 |
| 4 | 7 | 100 |
| 5 | 13 | 312 |
| 6 | 19 | 50 |
| 7 | 7 | 120 |
| 8 | 19 | 400 |
| 9 | 7 | 230 |
+------+----------+----------+
输出:
+----------+--------------------+
| name | travelled_distance |
+----------+--------------------+
| Elvis | 450 |
| Lee | 450 |
| Bob | 317 |
| Jonathan | 312 |
| Alex | 222 |
| Alice | 120 |
| Donald | 0 |
+----------+--------------------+
# Write your MySQL query statement below
select user.name name,IFNULL(sum(distance),0) travelled_distance
from Users user left join Rides r on r.user_id = user.id
group by user.id
order by travelled_distance desc,name
select sell_date,count(distinct product) num_sold,group_concat(distinct product) products from Activities
group by sell_date
GROUP_CONCAT
是 MySQL 中的聚合函数,用于将分组后的结果集中的多个行合并成一个字符串,并以指定的分隔符分隔它们。它的一般语法如下:GROUP_CONCAT(column_name [ORDER BY column_name] [SEPARATOR separator])
column_name
:要合并的列的名称。ORDER BY column_name
:可选,用于指定如何对合并的值进行排序。SEPARATOR separator
:可选,用于指定合并后的字符串中的分隔符。以下是一个示例,假设我们有一个名为 "students" 的表,其中包含学生的姓名,并且每个学生都属于一个特定的班级:
SELECT class_id, GROUP_CONCAT(student_name ORDER BY student_name SEPARATOR ',') AS students FROM students GROUP BY class_id;
查询患有 I 类糖尿病的患者 ID (patient_id)、患者姓名(patient_name)以及其患有的所有疾病代码(conditions)。I 类糖尿病的代码总是包含前缀 DIAB1 。
按 任意顺序 返回结果表。
查询结果格式如下示例所示。
示例 1:
输入:
Patients表:
+------------+--------------+--------------+
| patient_id | patient_name | conditions |
+------------+--------------+--------------+
| 1 | Daniel | YFEV COUGH |
| 2 | Alice | |
| 3 | Bob | DIAB100 MYOP |
| 4 | George | ACNE DIAB100 |
| 5 | Alain | DIAB201 |
+------------+--------------+--------------+
输出:
+------------+--------------+--------------+
| patient_id | patient_name | conditions |
+------------+--------------+--------------+
| 3 | Bob | DIAB100 MYOP |
| 4 | George | ACNE DIAB100 |
+------------+--------------+--------------+
解释:Bob 和 George 都患有代码以 DIAB1 开头的疾病。
select * from Patients
where conditions like 'DIAB1%' or conditions like '% DIAB1%'
送分题
select Users.name as NAME,tem.BALANCE from (
select account, sum(amount) as BALANCE from Transactions as t
group by account
having sum(amount)>10000) as tem join Users on tem.account=Users.account
有一些顾客可能光顾了购物中心但没有进行交易。请你编写一个解决方案,来查找这些顾客的 ID ,以及他们只光顾不交易的次数。
返回以 任何顺序 排序的结果表。
返回结果格式如下例所示。
示例 1:
输入:
Visits
+----------+-------------+
| visit_id | customer_id |
+----------+-------------+
| 1 | 23 |
| 2 | 9 |
| 4 | 30 |
| 5 | 54 |
| 6 | 96 |
| 7 | 54 |
| 8 | 54 |
+----------+-------------+
Transactions
+----------------+----------+--------+
| transaction_id | visit_id | amount |
+----------------+----------+--------+
| 2 | 5 | 310 |
| 3 | 5 | 300 |
| 9 | 5 | 200 |
| 12 | 1 | 910 |
| 13 | 2 | 970 |
+----------------+----------+--------+
输出:
+-------------+----------------+
| customer_id | count_no_trans |
+-------------+----------------+
| 54 | 2 |
| 30 | 1 |
| 96 | 1 |
+-------------+----------------+
解释:
ID = 23 的顾客曾经逛过一次购物中心,并在 ID = 12 的访问期间进行了一笔交易。
ID = 9 的顾客曾经逛过一次购物中心,并在 ID = 13 的访问期间进行了一笔交易。
ID = 30 的顾客曾经去过购物中心,并且没有进行任何交易。
ID = 54 的顾客三度造访了购物中心。在 2 次访问中,他们没有进行任何交易,在 1 次访问中,他们进行了 3 次交易。
ID = 96 的顾客曾经去过购物中心,并且没有进行任何交易。
如我们所见,ID 为 30 和 96 的顾客一次没有进行任何交易就去了购物中心。顾客 54 也两次访问了购物中心并且没有进行任何交易。
select v.customer_id,count(*) as count_no_trans from Visits as v left join Transactions as t on v.visit_id=t.visit_id
where t.transaction_id is null
group by v.customer_id
select * from Users
where mail regexp '^[a-zA-Z]+[a-zA-Z0-9_\\.\\/\\-]*@leetcode\\.com$';
算数
select contest_id,round(100*count(user.user_name)/(select count(*) from Users),2) as percentage
from Register as re left join Users as user on re.user_id = user.user_id
group by re.contest_id
order by percentage desc,contest_id
作差,思路清奇
输入:
Activity table:
+------------+------------+---------------+-----------+
| machine_id | process_id | activity_type | timestamp |
+------------+------------+---------------+-----------+
| 0 | 0 | start | 0.712 |
| 0 | 0 | end | 1.520 |
| 0 | 1 | start | 3.140 |
| 0 | 1 | end | 4.120 |
| 1 | 0 | start | 0.550 |
| 1 | 0 | end | 1.550 |
| 1 | 1 | start | 0.430 |
| 1 | 1 | end | 1.420 |
| 2 | 0 | start | 4.100 |
| 2 | 0 | end | 4.512 |
| 2 | 1 | start | 2.500 |
| 2 | 1 | end | 5.000 |
+------------+------------+---------------+-----------+
输出:
+------------+-----------------+
| machine_id | processing_time |
+------------+-----------------+
| 0 | 0.894 |
| 1 | 0.995 |
| 2 | 1.456 |
+------------+-----------------+
解释:
一共有3台机器,每台机器运行着两个进程.
机器 0 的平均耗时: ((1.520 - 0.712) + (4.120 - 3.140)) / 2 = 0.894
机器 1 的平均耗时: ((1.550 - 0.550) + (1.420 - 0.430)) / 2 = 0.995
select machine_id,round(sum(if(activity_type='start',-timestamp,timestamp))/(0.5*count(process_id)),3) as processing_time from Activity
group by machine_id
大小写转换
# Write your MySQL query statement below
select user_id, concat(upper(left(name,1)),lower(right(name,length(name)-1))) as name from Users
order by user_id
送分题
select tweet_id from Tweets
where length(content)>15
好几个去重
其实没必要对日期去重,后面的已经去过了
# Write your MySQL query statement below
select distinct date_id,make_name,count(distinct lead_id) unique_leads,count(distinct partner_id) unique_partners from DailySales
group by date_id,make_name
送
select user_id,count(follower_id) as followers_count from Followers
group by user_id
order by user_id
对于此问题,我们将至少有一个其他员工需要向他汇报的员工,视为一个经理。
编写SQL查询需要听取汇报的所有经理的ID、名称、直接向该经理汇报的员工人数,以及这些员工的平均年龄,其中该平均年龄需要四舍五入到最接近的整数。
返回的结果集需要按照 employee_id 进行排序。
查询结果的格式如下:
Employees table:
+-------------+---------+------------+-----+
| employee_id | name | reports_to | age |
+-------------+---------+------------+-----+
| 9 | Hercy | null | 43 |
| 6 | Alice | 9 | 41 |
| 4 | Bob | 9 | 36 |
| 2 | Winston | null | 37 |
+-------------+---------+------------+-----+
Result table:
+-------------+-------+---------------+-------------+
| employee_id | name | reports_count | average_age |
+-------------+-------+---------------+-------------+
| 9 | Hercy | 2 | 39 |
+-------------+-------+---------------+-------------+
Hercy 有两个需要向他汇报的员工, 他们是 Alice and Bob. 他们的平均年龄是 (41+36)/2 = 38.5, 四舍五入的结果是 39.
select tem.id1 employee_id,tem.name1 name,count(tem.id2) reports_count,round(avg(tem.age2),0) average_age from
(select e1.employee_id id1,e1.name name1,e1.reports_to to1,e1.age age1,
e2.employee_id id2,e2.name name2,e2.reports_to to2,e2.age age2 from Employees as e1 left join Employees as e2 on e1.employee_id=e2.reports_to
where e2.employee_id is not null) tem
group by tem.id1
order by employee_id
送分
select event_day day, emp_id,sum(out_time-in_time) total_time from Employees
group by event_day,emp_id
送分
select product_id from Products
where low_fats='Y' and recyclable='Y'
select employee_id, min(department_id) department_id from Employee
group by employee_id
having count(primary_flag) = 1
union
select employee_id, department_id from Employee
where primary_flag = 'Y'
请你重构 Products 表,查询每个产品在不同商店的价格,使得输出的格式变为(product_id, store, price) 。如果这一产品在商店里没有出售,则不输出这一行。
输出结果表中的 顺序不作要求 。
查询输出格式请参考下面示例。
示例 1:
输入:
Products table:
+------------+--------+--------+--------+
| product_id | store1 | store2 | store3 |
+------------+--------+--------+--------+
| 0 | 95 | 100 | 105 |
| 1 | 70 | null | 80 |
+------------+--------+--------+--------+
输出:
+------------+--------+-------+
| product_id | store | price |
+------------+--------+-------+
| 0 | store1 | 95 |
| 0 | store2 | 100 |
| 0 | store3 | 105 |
| 1 | store1 | 70 |
| 1 | store3 | 80 |
+------------+--------+-------+
解释:
产品 0 在 store1、store2、store3 的价格分别为 95、100、105。
产品 1 在 store1、store3 的价格分别为 70、80。在 store2 无法买到。
行列转换
# Write your MySQL query statement below
select product_id, 'store1' as store, store1 as price from Products where store1 is not null
union all
select product_id, 'store2' as store, store2 as price from Products where store2 is not null
union all
select product_id, 'store3' as store, store3 as price from Products where store3 is not null
select employee_id,if(employee_id%2=1,if(left(name,1)!='M',salary,0),0) bonus from Employees
order by employee_id
送
select user_id ,max(time_stamp) last_stamp from Logins
where time_stamp>='2020-01-01 00:00:00' and time_stamp<='2020-12-31 23:59:59'
group by user_id
第七穿插连
select employee_id from Employees where employee_id not in (select employee_id from salaries)
union
select employee_id from Salaries where employee_id not in (select employee_id from Employees)
order by employee_id
查找这些员工的id,他们的薪水严格少于$30000 并且他们的上级经理已离职。当一个经理离开公司时,他们的信息需要从员工表中删除掉,但是表中的员工的manager_id 这一列还是设置的离职经理的id 。
返回的结果按照employee_id 从小到大排序。
查询结果如下所示:
示例:
输入:
Employees table:
+-------------+-----------+------------+--------+
| employee_id | name | manager_id | salary |
+-------------+-----------+------------+--------+
| 3 | Mila | 9 | 60301 |
| 12 | Antonella | null | 31000 |
| 13 | Emery | null | 67084 |
| 1 | Kalel | 11 | 21241 |
| 9 | Mikaela | null | 50937 |
| 11 | Joziah | 6 | 28485 |
+-------------+-----------+------------+--------+
输出:
+-------------+
| employee_id |
+-------------+
| 11 |
+-------------+
解释:
薪水少于 30000 美元的员工有 1 号(Kalel) 和 11号 (Joziah)。
Kalel 的上级经理是 11 号员工,他还在公司上班(他是 Joziah )。
Joziah 的上级经理是 6 号员工,他已经离职,因为员工表里面已经没有 6 号员工的信息了,它被删除了。
表述不清
select e.employee_id
from Employees e
left join Employees ee
on e.manager_id = ee.employee_id
where e.manager_id is not null and ee.employee_id is null and e.salary < 30000
order by e.employee_id
送
select teacher_id,count(distinct subject_id) cnt from Teacher
group by teacher_id