简单难度

175. 组合两个表（left join … on）

题目描述


为避免数据丢失，故使用外连接，考虑到地址信息表可能有人没有填写（即地址表中没有该人员信息）。所以用Person 表 左外连接 Adress 表。

select FirstName,LastName,City,State
from Person as t1
left join Address as t2
on t1.PersonId = t2.PersonId;

参考的题解链接

176*. 第二高的薪水（ifnull ; ）类似题目177

题目：

题解:需要关注的点有两个，一个是ifnull 的位置，还有无论如何都要有select

select ifnull((select Salary from Employeegroup by Salaryorder by Salary desclimit 1,1),null) as `SecondHighestSalary`; # limit offset,count

方法二：窗函数

selectifnull((select distinct Salary  from (select Salary,dense_rank() over( order by Salary desc) as rn from Employee)as tempwhere rn = 2
),null) as `SecondHighestSalary`;

参考题解

181. 超过经理收入的员工（判断是否为空（is null）; 自联表）

方法一：（自己边分析边写的，多此一举了，自联表更快，请移步方法二）：
t1表：

select Name,Salary,ManagerId from Employee where ManagerId is not null;

Name	Salary	ManagerId
Joe	70000	3
Henry	80000	4

最终：

select t1.Name as `Employee` from
(select Name,Salary,ManagerId from Employee where ManagerId is not null) t1 , Employee t2
where t1.ManagerId=t2.Id and t1.Salary > t2.Salary;

方法二：（自联表）

select t1.Name as `Employee` from
Employee t1 , Employee t2
where t1.ManagerId=t2.Id and t1.Salary > t2.Salary;

方法三：使用 join 连表

select t1.Name as `Employee` from
Employee t1 join Employee t2
on
t1.ManagerId=t2.Id and t1.Salary > t2.Salary;

182.查找重复的电子邮箱（group by）

方法一：(group by)
这道题比较有意思的地方是，求的是重复出现过的邮箱，需要一个中间表来记录 Email,Email 对应出现的次数.

select Email from (select Email,count(Email) as num from Person group by Email) temp
where num != 1;

方法二：（group by + having）重要*

select Email from Person group by Email having count(Email)>1;

183. *从不订购的客户（子查询，not in）

题目描述：

法一：（子查询 + not in ）

select t1.Name as `Customers` from Customers as t1
where t1.Id not in (select CustomerId from Orders
);

法二：（左外连接）

select Name as `Customers`
from Customers t1
left join Orders t2
on t1.Id = t2.CustomerId
where t2.CustomerId is null;

196. 删除重复的电子邮箱 （delete , 自联表）

delete p1from Person p1,Person p2where p1.Email = p2.Email and p1.Id > p2.Id;

197*. 上升的温度(cross join ; datediff() )

题目解析 及 cross join ;datediff 介绍

题目介绍：

题解：

select t1.id
from Weather as t1 cross join Weather as t2 on datediff(t1.recordDate,t2.recordDate)=1
where t1.temperature > t2.temperature;

题解2：cross join 只是生成笛卡尔集，使用join也可以
cross join ; join ; inner join 区别

select t1.id
from Weather t1 join Weather t2 on datediff(t1.recordDate,t2.recordDate)=1
where t1.Temperature > t2.Temperature;
# datediff 的值是前 - 后

595. 大的国家（弱智题，没营养的）

题目：

题解：

select name,population,area from World
where area>3000000 or population>25000000;

596. (group by ,having ; 子查询)

题目描述：

题目解析链接
方法一：group by + having

select class
from courses
group by class
having count(distinct student)>=5;

方法二：子查询
先根据课程分组，查询课程及该课程的学生人数（去重后）作为临时表。再查询课程名称，条件是人数大于或者等于5

select class from(select class,count(distinct student) as `count` from courses group by class
) as temp
where `count`>=5;

620. 有趣的电影（按位与(&)|mod(num,2) 判断奇偶；order by 默认升序，且只能放在后边）

题目链接

select * from cinema
where id&1 and description!='boring'
order by rating desc ;

mod(id,2) = 1 返回id号是奇数的id

select id,movie,description,rating
from cinema
where description!='boring' and mod(id,2)=1
order by rating desc;

627*. 变更性别（update set ; case when）

题目描述：

题解： 注意update set的使用

update salary
set
sex =case sexwhen 'm' then 'f'else 'm'end;

update 使用方法链接

1179. 重新格式化部门表（group by 与 sum() ; case when; ）

mysql case when 用法可以看这里

select id,
sum(case when month='Jan' then revenue end )as Jan_Revenue,
sum(case when month='Feb' then revenue end )as Feb_Revenue,
sum(case when month='Mar' then revenue end )as Mar_Revenue,
sum(case when month='Apr' then revenue end )as Apr_Revenue,
sum(case when month='May' then revenue end )as May_Revenue,
sum(case when month='Jun' then revenue end )as Jun_Revenue,
sum(case when month='Jul' then revenue end )as Jul_Revenue,
sum(case when month='Aug' then revenue end )as Aug_Revenue,
sum(case when month='Sep' then revenue end )as Sep_Revenue,
sum(case when month='Oct' then revenue end )as Oct_Revenue,
sum(case when month='Nov' then revenue end )as Nov_Revenue,
sum(case when month='Dec' then revenue end )as Dec_Revenue
from Department
group by id
order by id;

理解：为什么使用sum聚合函数：

也就是说如果不使用聚合函数而是直接查询，那么第一组的数据（id=1）只有第一行会被检索到，后面两行不会被检索到，会使结果变为空值。
即：

id	Jan_Revenue	Feb_Revenue	Mar_Revenue	…	Dec_Revenue
1	8000	null	null	…	null
2	9000	null	null	…	null
3	null	10000	null	…	null

而非正确答案

id	Jan_Revenue	Feb_Revenue	Mar_Revenue	…	Dec_Revenue
1	8000	7000	6000	…	null
2	9000	null	null	…	null
3	null	10000	null	…	null

中等难度:

177*. 第N高的薪水(mysql函数 ；窗口函数；)

题目描述：

解法一（窗函数的使用）：

CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGINRETURN (# Write your MySQL query statement below.select distinct Salary from (select Salary,dense_rank() over( order by Salary desc) as rn from Employee)as tempwhere rn = N);
END

解法二 （单表查询，group by + order by + limit）

题解：注意赋值语句后面要加封号的

CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGINset N:= N-1;RETURN (# Write your MySQL query statement below.select Salaryfrom Employeegroup by Salaryorder by Salary desclimit N,1);
END

limit & offset 用法参考

其他解法

六种解法链接

178. 分数排名(窗函数经典问题)

题目描述：

解析：

注意Rank 是关键字，作为变量需要引起来 `Rank`

select Score,dense_rank() over (order by Score desc) as `Rank` from Scores;

参考答案

180*. 连续出现的数字（自联表 / 窗函数*(连续出现N次问题)）

解析链接
使用自联表：

select distinct t1.num as ConsecutiveNums from
Logs as t1,
Logs as t2,
Logs as t3
where t1.id = t2.id-1and t2.id = t3.id-1and t1.num = t2.numand t2.num = t3.num;

使用窗函数

考查窗口函数lag、lead的用法:(*重要)

知识点链接

使用lead()

select distinct num as `ConsecutiveNums`
from(select num,lead(num,1) over(order by id) as num2,lead(num,2) over(order by id) as num3from Logs
) as t
where t.num = t.num2 and t.num = t.num3;

使用lag()

select distinct num as `ConsecutiveNums`
from(select num,lag(num,1) over(order by id) as num2,lag(num,2) over(order by id) as num3from Logs
) as t
where t.num = t.num2 and t.num = t.num3;

184. 部门工资最高的员工（窗函数）–具体解析看185

使用窗函数解法：

select dname as `Department`,ename as `Employee`,Salary
from (select t1.Name as ename,Salary,t2.Name as dname, dense_rank() over (partition by DepartmentId order by Salary desc) as rn from Employee as t1, Department as t2 where t1.DepartmentId = t2.Id
) as temp
where rn <=1;

非窗函数解法：

select t3.Name as `Department`,temp.Name as `Employee`,Salary from
(select Name,t1.Salary,DepartmentId from Employee as t1where 1 > (select count(distinct Salary) from Employee as t2where t1.Salary < t2.Salary and t1.DepartmentId = t2.DepartmentId)
) as temp, Department as t3 where temp.DepartmentId = t3.Id;

626. 换座位(case when;可以 from表和查询出来的值)

分布题解：

首先计算座位数，因为后边要判断奇偶

select count(*) as counts from seat;

接下来搜索 id 和 学生 ,需要注意一点case when 语句可以引用counts ,但是引用seat_counts时会报错

select
(casewhen id!= counts and mod(id,2)!=0 then id+1when id = counts and mod(id,2)!=0 then idelseid-1
end) as id, student
from seat,(select count(*) as counts from seat) as seat_counts;

id	student
2	Abbot
1	Doris
4	Emerson
3	Green
5	Jeames

最终：使用order by 重新调整顺序

select
(casewhen id!= counts and mod(id,2)!=0 then id+1when id = counts and mod(id,2)!=0 then idelseid-1
end) as id, student
from seat,(select count(*) as counts from seat) as seat_counts
order by id;

困难难度

185. 部门工资前三高的员工（窗函数）

使用窗函数查询

使用窗函数：

先使用窗函数求出分部门并以工资排序的新表，再利用该中间表做查询

select t2.Name as dname,t1.Name as ename,Salary,dense_rank() over(partition by DepartmentId order by Salary desc) as rn
from Employee as t1, Department as t2 where t1.DepartmentId=t2.Id;

dname	ename	Salary	rn
IT	Max	90000	1
IT	Joe	85000	2
IT	Randy	85000	2
IT	Will	70000	3
IT	Janet	69000	4
Sales	Henry	80000	1
Sales	Sam	60000	2

最终查询：

select dname as `Department` ,ename as `Employee`,Salary from
(select t2.Name as dname,t1.Name as ename,Salary,dense_rank() over(partition by DepartmentId order by Salary desc) as rn
from Employee as t1, Department as t2 where t1.DepartmentId=t2.Id) temp
where rn <=3;

Department	Employee	Salary
IT	Max	90000
IT	Randy	85000
IT	Joe	85000
IT	Will	70000
Sales	Henry	80000
Sales	Sam	60000

补充：rank(),dense_rank(),row_num()

不使用窗函数查询 (自连表)

先自连表查出薪资前三的姓名，工资，部门编号作为新表

select Name,t1.Salary,DepartmentId from Employee as t1
where 3 > (select count(distinct Salary) from Employee as t2where t1.Salary < t2.Salary and t1.DepartmentId = t2.DepartmentId
);

Name	Salary	DepartmentId
Joe	85000	1
Henry	80000	2
Sam	60000	2
Max	90000	1
Randy	85000	1
Will	70000	1

select t3.Name as `Department`, temp.Name as `Employee`, Salary
from
(select Name,t1.Salary,DepartmentId from Employee as t1where 3 > (select count(distinct Salary) from Employee as t2where t1.Salary < t2.Salary and t1.DepartmentId = t2.DepartmentId)
) as temp, Department as t3
where temp.DepartmentId = t3.Id;

Department	Employee	Salary
IT	Joe	85000
Sales	Henry	80000
Sales	Sam	60000
IT	Max	90000
IT	Randy	85000
IT	Will	7000

262 .行程和用户*

题目：




题目解析：



解法：

select request_at as `Day`,
round(sum(if(t.status = 'completed',0,1))/count(t.status),2) as `Cancellation Rate`
from Trips as t
join Users as u1 on (t.client_id = u1.users_id and u1.banned = "No")
join Users as u2 on (t.driver_id = u2.users_id and u2.banned = "No")
where request_at between "2013-10-01" and "2013-10-03"
group by request_at;

其他解法：这里

601.体育馆的人流量

题目：

方法一：自联表

题目解析

select distinct s1.* from Stadium s1,Stadium s2, Stadium s3
where s1.people >= 100 and s2.people >= 100 and s3.people >= 100
and ((s1.id + 1 = s2.id and s1.id + 2 = s3.id and s2.id + 1 = s3.id) or  -- s1是巅峰期第一天(s1.id - 1 = s2.id and s1.id + 1 = s3.id and s2.id + 2 = s3.id) or  -- s1是巅峰期第二天(s1.id - 2 = s2.id and s1.id - 1 = s3.id and s2.id + 1 = s3.id)     -- s1是巅峰期第三天
)
order by id;

方法二：（待补充）

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