检测 SQL 中的连续日期并启动计数器

Question

Dates
2021-01-01
2021-01-02
2021-01-03
2021-01-10
2021-01-11
2021-01-12
2021-01-15

预期输出：

Dates	Counter
2021-01-01	1
2021-01-02	2
2021-01-03	3
2021-01-10	1
2021-01-11	2
2021-01-12	3
2021-01-15	1

知道如何实现吗？我确实知道如何找到连续日期的最小值、最大值和计数，但需要一个计数器，每次出现日期间隔时都会重置

Answer 1

解决此类“缺口和孤岛”问题的一种方法是根据与前一天的差异计算排名。

然而，计算日期之间天数差异的方法取决于您的 RDBMS。

此示例使用 DATEDIFF（MS Sql 服务器）

SELECT 
 Dates, 
 ROW_NUMBER() OVER (PARTITION BY Rnk ORDER BY Dates) AS Counter
FROM
(
    SELECT 
     Dates, 
     SUM(CASE 
         WHEN DATEDIFF(day, prevDate, Dates) = 1 THEN 0 
         ELSE 1 END) OVER (ORDER BY Dates) AS Rnk 
    FROM
    (
      SELECT 
       Dates, 
       LAG(Dates) OVER (ORDER BY Dates) AS prevDate
      FROM your_table
    ) q1
) q2
ORDER BY Dates;

Answer 2

你可以使用自join:

with cte as (
   select row_number() over (order by dt) r, dt from test
)
select c.dt, sum(c.dt = date_add(c1.dt, interval c.r-c1.r day)) 
from cte c join cte c1 on c1.r <= c.r group by c.dt order by c.dt

检测 SQL 中的连续日期并启动计数器

Detect consecutive dates in SQL and start counter

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gaps-and-islands