SQL——连续出现的数字

1. SQL三个排序函数 ROW_NUMBER()、RANK()、DENSE_RANK()

ROW_NUMBER()不并列 连续的
RANK()分组不连续排序（跳跃排序）
DENSE_RANK()并列连续

创建实例表：

DROP table IF EXISTS con;
Create Table IF NOT EXISTS con(id int,Num int);
INSERT INTO con VALUES(1,1);
INSERT INTO con VALUES(2,1);
INSERT INTO con VALUES(3,1);
INSERT INTO con VALUES(4,4);
INSERT INTO con VALUES(5,2);
INSERT INTO con VALUES(6,2);
INSERT INTO con VALUES(7,3);
INSERT INTO con VALUES(8,3);

实例表：

三个函数 ROW_NUMBER()、 RANK()、 DENSE_RANK() 的区别如下图：

SELECT *,
	ROW_NUMBER() over(ORDER BY Num DESC) AS 'ROW_NUMBER()',
	RANK() over(ORDER BY Num DESC) AS 'RANK()',
	DENSE_RANK() over(ORDER BY Num DESC) AS 'DENSE_RANK()'
FROM con;

函数区别结果：

2. TOP N 问题

返回第二高的数字

-- 方法一：DENSE_RANK() 窗口函数
SELECT DISTINCT Num
FROM (
	SELECT *,
		DENSE_RANK() over(ORDER BY Num) AS dense
	FROM con
)  AS a
WHERE dense = 2;

-- 方法二： 相关子查询
SELECT DISTINCT Num
FROM con AS c1
WHERE 2 = (
	SELECT COUNT(DISTINCT Num)
	FROM con AS c2
	WHERE c1.Num >= c2.Num
)

结果如下：

3. 返回连续出现 N 次的数

返回连续出现三次的数字

• 情况一：要求 id 连续，而不是表格中的原顺序 的情况下Num连续

-- 方法一： join on
SELECT DISTINCT c1.Num
FROM con AS c1
JOIN con AS c2 ON c1.Num = c2.Num AND c2.id = c1.id   1 
JOIN con AS c3 ON c3.Num = c2.Num AND c3.id = c2.id   1 
-- 方法二：in
SELECT DISTINCT Num
SELECT *
FROM con 
WHERE (id   1, Num) IN (SELECT * FROM con)
	AND (id   2, Num) IN (SELECT * FROM con)

结果如下：

• 情况二：按 表格中的行顺序 Num连续

-- 方法一：排序函数
SELECT DISTINCT Num 
FROM (
	SELECT *,
-- 	id 索引值
	ROW_NUMBER() over(ORDER BY id) AS number,
-- 	分组排序
	DENSE_RANK() over(PARTITION BY Num ORDER BY id) AS dense,
-- 	差值
	ROW_NUMBER() over(ORDER BY id) - DENSE_RANK() over(PARTITION BY Num ORDER BY id) AS different
FROM con
) AS a
GROUP BY Num, different
HAVING count(*) >= 3
-- 方法二： 偏移函数
-- lead（）向上偏移

SELECT DISTINCT Num
FROM (
	SELECT *,
-- 	 按照表格中的原顺序， num连续出现相同值
		LEAD(Num, 1) over(ORDER BY id) AS l1,
		LEAD(Num,2) over(ORDER BY id) AS l2
	FROM con
) AS a
WHERE Num = l1 AND l1 = l2;
-- lag() 向下偏移
SELECT DISTINCT Num
FROM (
	SELECT *, 
	-- 	按照表格中的原顺序， num连续出现相同值
		LAG(Num, 1) over(ORDER BY id) AS l1,
		LAG(Num, 2) over(ORDER BY id) AS l2
	FROM con
) AS a
WHERE Num = l1 AND l1 = l2;

结果如下：

总结：

1. TOP N 问题：排序函数 和 相关子查询方法
2. 返回连续出现数字
   ①连续指的是 按表格中顺序连续出现，排序函数和偏移函数
   ②id连续，JOIN ON 和 IN 两种方法