Mysql刷题

1.第二高的薪水

要求：如果有就查出来。若无则null

    SELECT IFNULL(
	(SELECT DISTINCT salary FROM Employee ORDER BY  salary DESC LIMIT 1,1),NULL) AS SecondHighestSalary

思路：采用分页思想，用到了LIMIT函数和IFNULL函数
- LIMIT m,n从m+1开始取n个
- IFNULL(expression_1,expression_2) 如果expression_1为空就返回expression_2否则返回expression_1
- LIMIT m OFFSET n 表示跳过n个数据读取m条数据

2.第N高的薪水

思路：和第一题思路相同

CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGIN
  #如果不减一，取得就是第n+1高的
  set N=N-1;
  RETURN (
      # Write your MySQL query statement below.
      SELECT IFNULL(
	(SELECT DISTINCT salary FROM Employee ORDER BY  salary DESC LIMIT 1 OFFSET N),NULL) AS SecondHighestSalary
  );
END

3.分数排名

要求：分数相同，并列名次，效果如下

Score	Rank
4.00	1
4.00	1
3.00	2
2.00	3

SELECT a.Score AS Score,b.Rank 
FROM Scores a  
        LEFT JOIN
	(
	    SELECT  s.Score,@rank:=@rank+1 AS Rank from (
			SELECT DISTINCT Score FROM Scores ORDER BY Score DESC
	    ) s,(SELECT @rank:=0) r
	) b
    ON a.Score=b.Score
ORDER BY a.Score DESC

思路：先去重并排名，再用左连接去关联
- 定义变量
  - 直接set @rank=0或者set @rank=0这里用不用“:”都可以
  - 查询结果若被用为子查询则不可用set，在from后定义（SELECT @rank:=0），在查询中必须要带“:”，不带表示的是一个比较语句

4.连续出现的数字

要求：查找所有至少连续出现三次的数字

Id	Num
1	1
2	1
3	1
4	2
5	1
6	2

select distinct Num as ConsecutiveNums
from (
  select Num, 
    case 
      when @prev = Num then @count := @count + 1
      when (@prev := Num) is not null then @count := 1
    end as CNT
  from Logs, (select @prev := null,@count := null) as t
) as temp
where temp.CNT >= 3

case...when...then end结束

5.超过经理收入的员工

要求：Employee表包含所有员工，他们的经理也属于员工。每个员工都有一个 Id，此外还有一列对应员工的经理的 Id。

Id	Name	Salary	ManagerId
1	Joe	70000	3
2	Henry	80000	4
3	Sam	60000	NULL
4	Max	90000	NULL

SELECT a.Name as Employee  FROM (
	SELECT a.Name,a.Salary,b.Salary as SalaryB
		from Employee a
			LEFT JOIN  Employee b
			on a.ManagerId=b.id
	WHERE a.ManagerId IS NOT NULL
) a
WHERE a.Salary>a.SalaryB

思路：利用左连接将数据查成如下格式，再比较一下即可（SalaryB为所属经理的薪水）

Name	Salary	SalaryB
Joe	70000	60000
Henry	80000	90000

6.部门工资前三高的所有员工

思路：可以使用变量先排名在关联如第3题，此处提供一种全新解法

Id	Name	Salary	DepartmentId
1	Joe	85000	1
2	Henry	80000	2
3	Sam	60000	2
4	Max	90000	1
5	Janet	69000	1
6	Randy	85000	1
7	Will	70000	1

SELECT e1.Salary,e1.name,e1.Department
FROM Employee AS e1
    WHERE 3 > 
	(SELECT  count(DISTINCT e2.Salary) 
	FROM	Employee AS e2 
	WHERE	e1.Salary < e2.Salary 	AND e1.Department = e2.Department) 
ORDER BY e1.department,e1.salary desc

举个例子：
- 假设e1 = e2 = [4,5,6,7,8]
- e1.Salary = 4，e2.Salary 可以取值 [5,6,7,8]，count(DISTINCT e2.Salary) = 4
- e1.Salary = 5，e2.Salary 可以取值 [6,7,8]，count(DISTINCT e2.Salary) = 3
- e1.Salary = 6，e2.Salary 可以取值 [7,8]，count(DISTINCT e2.Salary) = 2
- e1.Salary = 7，e2.Salary 可以取值 [8]，count(DISTINCT e2.Salary) = 1
- e1.Salary = 8，e2.Salary 可以取值 []，count(DISTINCT e2.Salary) = 0
所以 3>count(DISTINCT e2.Salary) 时，e1.Salary 的取值为[6,7,8]，即排名前三高的薪水

7.上升的温度

要求：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。

SELECT b.Id
FROM Weather as a,Weather as b
WHERE a.Temperature < b.Temperature and DATEDIFF(a.RecordDate,b.RecordDate) = -1;

思路：本题使用了DATEDIFF函数来计算两个日期的时间差。

+ DATEDIFF('2007-12-31','2007-12-30');   # 1
+ DATEDIFF('2010-12-30','2010-12-31');   # -1