184. Department Highest Salary
题意：
The Employee table holds all employees. Every employee has an Id, a salary, and there is also a column for the department Id. +----+-------+--------+--------------+ | Id | Name | Salary | DepartmentId | +----+-------+--------+--------------+ | 1 | Joe | 70000 | 1 | | 2 | Henry | 80000 | 2 | | 3 | Sam | 60000 | 2 | | 4 | Max | 90000 | 1 | +----+-------+--------+--------------+ The Department table holds all departments of the company. +----+----------+ | Id | Name | +----+----------+ | 1 | IT | | 2 | Sales | +----+----------+ Write a SQL query to find employees who have the highest salary in each of the departments. For the above tables, Max has the highest salary in the IT department and Henry has the highest salary in the Sales department. +------------+----------+--------+ | Department | Employee | Salary | +------------+----------+--------+ | IT | Max | 90000 | | Sales | Henry | 80000 | +------------+----------+--------+
解法：
1 select d.Name as Department, e.Name as Employee, e.Salary from Employee as e ,(select DepartmentId, max(Salary) max from Employee group by DepartmentId) t,Department as d where e.Salary = t.max and e.DepartmentId = t.DepartmentId and d.Id = e.DepartmentId;
176. Second Highest Salary
题意：
Write a SQL query to get the second highest salary from the Employee table. +----+--------+ | Id | Salary | +----+--------+ | 1 | 100 | | 2 | 200 | | 3 | 300 | +----+--------+ For example, given the above Employee table, the second highest salary is 200. If there is no second highest salary, then the query should return null.
解法：Using max() will return a NULL if the value doesn't exist. So there is no need to UNION a NULL. Of course, if the second highest value is guaranteed to exist, using LIMIT 1,1 will be the best answer.
1 select max(Salary) as SecondHighestSalary from Employee where Salary < (select max(Salary) from Employee);
197. Rising Temperature
题意：
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 | +----+ Subscribe to see which companies asked this question.
解法：
1 select a.Id as Id from Weather a, Weather b where to_days(a.Date)-to_days(b.Date) = 1 and a.Temperature > b.Temperature;