如何找到连续出现两次以上的号码？

Question

来源table:

id     num
-------------------
1       1
2       1
3       1
4       2
5       2
6       1

输出：（至少出现2次）

num   times
--------------
1      3
2      2

Answer 1

我会用间隙和孤岛技术来解决这个问题：

select num, max(cnt)
from (
    select num, count(*) cnt
    from (
     select 
          id, 
          num, 
          row_number() over(order by id) rn1,
          row_number() over(partition by num order by id) rn2
      from mytable
    ) t
    group by num, rn1 - rn2
) t
group by num

最内层查询计算整个 table 和 num 组内的行号；行号之间的差异为您提供了每条记录所属的一组相邻记录（您可以运行独立地进行子查询并了解差异如何演变以了解更多信息）。

然后，下一级统计每组相邻记录的记录数。最外层查询为每个 num.

获取最大相邻记录数

Demo on DB Fiddle:

num | (No column name)
--: | ---------------:
  1 |                3
  2 |                2

Answer 2

根据评论中定义的添加逻辑，这就是您所追求的：

WITH YourTable AS(
    SELECT V.id,
           V.num
    FROM (VALUES(1,1),
                (2,1),
                (3,1),
                (4,2),
                (5,2),
                (6,1),
                (7,1))V(id,num)), --Added extra row due to logic defined in comments
Grps AS(
    SELECT YT.id,
           YT.num,
           ROW_NUMBER() OVER (ORDER BY id) -
           ROW_NUMBER() OVER (PARTITION BY Num ORDER BY id) AS Grp
    FROM YourTable YT),
Counts AS(
    SELECT num,
           COUNT(num) AS Times
    FROM grps
    GROUP BY grp,
             num)
SELECT num,
       MAX(times) AS times
FROM Counts
GROUP BY num;

这使用 CTE 和 ROW_NUMBER 来定义组，然后使用额外的 CTE 来获得每个组的 COUNT。最后你可以获得 MAX COUNT 每个 num.

Answer 3

this will work for you

select num,count(num) times from Tabl
group by num

如何找到连续出现两次以上的号码？

how to find the number has more than two consecutive appearences?

sql

tsql

sql-server

window-functions

gaps-and-islands