Mysql习题
编写一个 SQL 查询,获取 Employee 表中第二高的薪水(Salary) 。
+----+--------+
| Id | Salary |
+----+--------+
| 1 | 100 |
| 2 | 200 |
| 3 | 300 |
+----+--------+
例如上述 Employee 表,SQL查询应该返回 200 作为第二高的薪水。如果不存在第二高的薪水,那么查询应返回 null。
+---------------------+
| SecondHighestSalary |
+---------------------+
| 200 |
+---------------------+
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/second-highest-salary
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
ifnull(A,B)语句,如果A不为空,则返回A,否则返回B。
limit x,y 从返回的第x+1条语句开始,返回y条
select ifnull((select distinct Salary from Employee order by Salary desc limit 1,1),null) as SecondHighestSalary;
给定一个 Weather 表,编写一个 SQL 查询,来查找与之前(昨天的)日期相比温度更高的所有日期的 Id。
+---------+------------------+------------------+
| Id(INT) | RecordDate(DATE) | Temperature(INT) |
+---------+------------------+------------------+
| 1 | 2015-01-01 | 10 |
| 2 | 2015-01-02 | 25 |
| 3 | 2015-01-03 | 20 |
| 4 | 2015-01-04 | 30 |
+---------+------------------+------------------+
例如,根据上述给定的 Weather 表格,返回如下 Id:
+----+
| Id |
+----+
| 2 |
| 4 |
+----+
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/rising-temperature
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
1 # Write your MySQL query statement below 2 select a.Id from Weather as a, Weather as b where a.Temperature > b.Temperature && datediff(a.RecordDate, b.RecordDate) = 1;