Question

+-----+------------+------------+---------------------+
| id  | seller_id  | prod_id    | date                |
+-----+------------+----------------------------------+
|   1 | 283        | 4243       | 2016-10-10 23:55:01 |
|   2 | 287        | 4243       | 2016-10-10 02:01:06 |
|   3 | 283        | 4243       | 2016-10-11 23:55:06 |
|   4 | 311        | 4243       | 2016-10-11 23:55:07 |
|   5 | 283        | 4243       | 2016-10-12 23:55:07 |
|   6 | 283        | 4243       | 2016-10-13 23:55:07 |
|   7 | 311        | 4243       | 2016-10-13 23:55:07 |
|   8 | 287        | 4243       | 2016-10-14 23:57:06 |
|   9 | 311        | 4243       | 2016-10-14 23:57:06 |
|  10 | 311        | 4243       | 2016-10-15 23:57:06 |
+-----+------------+------------+---------------------+

从上表中我将如何使用MySQL查询提取以下信息？

+------------+---------+----------------+---------------+
| seller_id  | prod_id | streak in days | begin streak  | 
+-----+------------+--------------------+---------------+
| 283        | 4243    | 4              | 2016-10-10    |
| 287        | 4243    | 1              | 2016-10-10    |
| 311        | 4243    | 1              | 2016-10-11    |
| 311        | 4243    | 3              | 2016-10-13    |
| 287        | 4243    | 1              | 2016-10-14    |
+------------+---------+----------------+---------------|

所以基本上我需要为销售产品的每个卖家（seller_id）确定每个连续日期的块（prod_id）。

我将此示例限制为1个prod_id且仅限几天，但卖家确实销售超过1个产品（prod_id）

Answer 1

SELECT
  seller_id
  ,prod_id
  ,COUNT(*) as StreakInDays
  ,MIN(DateCol) as BeginStreak
FROM
  (
    SELECT
      seller_id
      ,prod_id
      ,DATE(DateCol) as DateCol
      ,(@rn:= if((@seller = seller_id) AND (@prod = prod_id), @rn + 1,
                 if((@seller:= seller_id) AND (@prod:= prod_id), 1, 1)
                  )
        ) as RowNumber
    FROM
      Transact t
      CROSS JOIN (SELECT @seller:=0, @prod:=0, @rn:=0) var
    ORDER BY
      seller_id
      ,prod_id
      ,DATE(DateCol)
    ) t    
GROUP BY
  seller_id
  ,prod_id
  ,DATE_SUB(DateCol, INTERVAL RowNumber Day)
ORDER BY
  prod_id
  ,DATE_SUB(DateCol, INTERVAL RowNumber Day)
  ,seller_id

生成由seller_id和prod_id分区的分区行号。然后使用Date - RownNumber作为分组，您可以通过简单的聚合得到答案。

SQL小提示向您展示它适用于多种产品，卖家等。http://sqlfiddle.com/#!9/0a0c44/8/0

请注意，如果同一卖家可能在同一天有一个以上的产品交易，那么您需要在生成之前将Transact替换为DISTINCT seller_id，prod_id，DATE（日期）的派生表行号如下：

SELECT
  seller_id
  ,prod_id
  ,COUNT(*) as StreakInDays
  ,MIN(DateCol) as BeginStreak
FROM
  (
    SELECT
      seller_id
      ,prod_id
      ,DateCol
      ,(@rn:= if((@seller = seller_id) AND (@prod = prod_id), @rn + 1,
                 if((@seller:= seller_id) AND (@prod:= prod_id), 1, 1)
                  )
        ) as RowNumber
    FROM
     (SELECT DISTINCT seller_id, prod_id, DATE(DateCol) as DateCol
       FROM
         Transact
      )t
      CROSS JOIN (SELECT @seller:=0, @prod:=0, @rn:=0) var
    ORDER BY
      seller_id
      ,prod_id
      ,DateCol
    ) t    
GROUP BY
  seller_id
  ,prod_id
  ,DATE_SUB(DateCol, INTERVAL RowNumber Day)
ORDER BY
  prod_id
  ,DATE_SUB(DateCol, INTERVAL RowNumber Day)
  ,seller_id

http://sqlfiddle.com/#!9/0a0c44/11

MySQL计数/跟踪条纹或连续日期

1 个答案: