leetcode刷题 --数据库（更新中。。。）

组合两个表：

题目链接

• 解题思路：左连接

select FirstName, LastName, City, State 
from person natural left join address;
-- 或
select FirstName, LastName, City, State 
from person p left join address a on p.PersonID = a.PersonID

第二高的薪水：

题目链接

• 解题思路：用distinct去除重复值，倒序排序之后再用limit选出第二高的

select distinct Salary as SecondHighestSalary 
from Employee 
order by Salary desc 
limit 1,1;

无语，做错了，题目要求如果不存在第二高的薪水，那么查询应返回 null，这里用IFNULL()函数来实现

select IFNULL((select distinct Salary
from Employee
order by Salary desc
limit 1,1),null) as SecondHighestSalary

分数排名：

题目链接

• 解题思路：抄的，没想明白

select s1.Score, count(distinct(s2.score)) Rank
from Scores s1, Scores s2
where s1.score<=s2.score
group by s1.Id
order by Rank;

超过经理收入的员工：

题目链接

• 解题思路：写个子查询就完事了

select Name Employee from Employee e
where Salary > (select Salary 
from Employee where Id = e.Managerid);

查找重复的电子邮箱：

题目链接

• 解题思路：分组计数就完事了

select Email from person group by Email having count(Email)>1

从不订购的客户：

题目链接

• 解题思路：用个<> all就完事了

select Name as Customers 
from Customers 
where id <> all (select customerId from orders);

删除重复的电子邮箱：

题目链接

• 解题思路：删除重复的电子邮箱,并保存id最小的那个

delete p1 from person p1, person p2
where p1.Email = p2.Email and p1.Id > p2.Id;

上升的温度

题目链接

• 解题思路：利用datediff()函数来表示今天昨天

SELECT w1.Id 
FROM Weather w1, Weather w2 
WHERE
	DATEDIFF( w1.RecordDate, w2.RecordDate ) = 1 
	AND 
    w1.Temperature > w2.Temperature

大的国家

题目链接

• 解题思路：最简单的题了

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

超过5名学生的课：

题目链接

• 解题思路：也不难，就是要注意可能有重复数据

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

行程和用户：

题目链接

• 力扣说我错了，理由是：
  

select request_at as Day, 
round (
(sum(case when status = 3 then 1 else 0 end) 
+ sum( case when status = 2 then 1 else 0 end )
)/count(status),2)
as 'Cancellatio Rate'
from trips t join users u on t.Client_Id = u.Users_Id and u.banned = 'No' GROUP BY Request_at;