LeetCode有10道SQL的题目，最近学习SQL语言，顺便刷题强化一下，

说实话刷完SQL学习指南这本书，不是很难，上面的例子 跟语法规则我都能理解透，

实际中来做一些比较难的业务逻辑题，却一下子很难想起如何利用SQL来描述

还是要多刷题开阔自己的思路，把学到手的语法加以练习 应用到实际中去

197. Rising Temperature

Given a Weather table, write a SQL query to find all dates' Ids with higher temperature compared to its previous (yesterday's) dates.

+---------+------------+------------------+

| Id(INT) | Date(DATE) | Temperature(INT) |

+---------+------------+------------------+

|       1 | 2015-01-01 |               10 |

|       2 | 2015-01-02 |               25 |

|       3 | 2015-01-03 |               20 |

|       4 | 2015-01-04 |               30 |

+---------+------------+------------------+

For example, return the following Ids for the above Weather table:

+----+

| Id |

+----+

|  2 |

|  4 |

+----+

这个题目题意是要求找到比前一天气温高的条目

# Write your MySQL query statement below

select Id from Weather w1 where Temperature >(select Temperature from Weather w2 where to_days(w2.Date)=(to_days(w1.Date)-1))

这里用了子查询，针对外部查询的每一条记录，在子查询中找到前一天的气温 作为子查询的结果 最后与外部查询相比较，这样就能找出结果来

当然这里思路比较简单，我想出来的都是最简单 也是最容易理解的方案。

Write a SQL query to find all duplicate emails in a table named Person.

+----+---------+

| Id | Email   |

+----+---------+

| 1  | a@b.com |

| 2  | c@d.com |

| 3  | a@b.com |

+----+---------+

For example, your query should return the following for the above table:

+---------+

| Email   |

+---------+

| a@b.com |

+---------+

Note: All emails are in lowercase.

找出有完全一样email的条目，

解题的思路在于group by 通过集聚函数count计算条目，当条目大于2则显示，当然这里不能使用where 因为在SQL里面where必须在group by之前，然后才能集聚

所以此处只能使用having来完成 group by集聚完成之后的任务

select Email from Person group by Email having count(*)>=2

181. Employees Earning More Than Their Managers

找出比他们经理还挣得多的员工

毫无疑问子查询duang起

select Name Employee from Employee a where Salary >(select Salary from Employee where Id=a.ManagerId)

175. Combine Two Tables

这题直接外连，不用讲思路了

select FirstName, LastName, City, State from Person left join Address on Address.PersonId=Person.PersonId