如何查找连续登录的日期

Question

我的数据库中有 table (MySQL 5.7.36)，我试图找到连续的一天有条件

如果连续一天 > 7

连续一天将被设置为零

DATE_SERV
2022-01-01
2022-01-02
2022-01-03
2022-01-05
2022-01-06
2022-01-09
2022-01-10
2022-01-11

我的实际期望 table 是

DATE_SERV	day_consecutive
2022-01-01	1
2022-01-02	2
2022-01-03	3
2022-01-05	1
2022-01-06	2
2022-01-09	1
2022-01-10	2
2022-01-11	3
2022-01-12	4
2022-01-13	5
2022-01-14	6
2022-01-15	7
2022-01-16	1
2022-01-17	2

Answer 1

我之前写过这个，以为你在使用 MySQL 8.x（支持 window functions，不幸的是 5.x 不支持）。无论如何，只是张贴它以防它对其他人有用......

您可以采用此博客 Gaps and Islands Across Date Ranges 中的方法。首先确定“岛屿”或连续日期组

SELECT
        DATE_SERV
        , SUM( IF( DATEDIFF(DATE_SERV, Prev_Date) = 1, 0, 1) ) OVER(
                   ORDER BY DATE_SERV 
        )  AS DateGroup_Num
FROM
      (
         SELECT DATE_SERV
                , LAG(DATE_SERV,1) OVER (
                       ORDER BY DATE_SERV
                ) AS Prev_Date
         FROM   YourTable
      ) grp

产生这个结果：

DATE_SERV	DateGroup_Num
2022-01-01	1
2022-01-02	1
2022-01-03	1
2022-01-05	2
2022-01-06	2
2022-01-09	3
2022-01-10	3
2022-01-11	3

然后使用条件 SUM(...) 查找每组的最早日期，并显示自该日期以来的连续天数：

SELECT 
       t.DATE_SERV
       , DATEDIFF( 
            t.DATE_SERV
            , MIN(t.DATE_SERV) OVER( 
                  PARTITION BY t.DateGroup_Num 
                  ORDER BY t.DATE_SERV 
             )
        ) +1 AS Consecutive_Days
FROM ( 
        SELECT
               DATE_SERV
              , SUM( IF( DATEDIFF(DATE_SERV, Prev_Date) = 1, 0, 1) ) OVER(
                    ORDER BY DATE_SERV 
                )  AS DateGroup_Num
        FROM
        (
                SELECT DATE_SERV
                       , LAG(DATE_SERV,1) OVER (
                            ORDER BY DATE_SERV
                         ) AS Prev_Date
                FROM   YourTable
        ) grp
    ) t

结果：

DATE_SERV	Consecutive_Days
2022-01-01	1
2022-01-02	2
2022-01-03	3
2022-01-05	1
2022-01-06	2
2022-01-09	1
2022-01-10	2
2022-01-11	3

db<>fiddle here

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