按某列分组的移位操作SQL

Question

所以我和 GreenPlum 一起在一个大的 table 名称 purchases 上包含超过 400 万行。这是 table:

的示例

userId |        purchaseTime      | timeDiff
------------------------------------------
 17    |   2016-02-01 11:01:02    |
 17    |   2016-02-01 13:24:58    |
 17    |   2016-02-01 21:12:36    |
 67    |   2016-02-01 17:04:49    |
 84    |   2016-02-01 16:13:20    |
 94    |   2016-02-01 05:46:13    |
 94    |   2016-02-01 21:33:19    |

table 是按 userID 和 purchaseTime 排序的，以帮助理解我的目标

我的 objective 是通过包含每个用户当前行与上次购买时间之间的时间差来更新此 table。

让它看起来像这样：

userId |        purchaseTime      | timeDiff
------------------------------------------
 17    |   2016-02-01 11:01:02    | NULL
 17    |   2016-02-01 13:24:58    | 2:23:56
 17    |   2016-02-01 21:12:36    | 8:12:38
 67    |   2016-02-01 17:04:49    | NULL
 84    |   2016-02-01 16:13:20    | NULL
 94    |   2016-02-01 05:46:13    | NULL
 94    |   2016-02-01 21:33:19    | 16:13:06

您的回答中的 select 对我有帮助。现在我需要执行 UPDATE，但在执行 UPDATE 时出现语法错误：

WITH tmp_table AS
(
    SELECT userId ,  
       purchaseTime ,
       purchaseTime - LAG(purchaseTime )
       OVER (PARTITION BY userId  ORDER BY purchaseTime) AS timeDiff
    FROM   purchases
)

UPDATE purchases SET timeDiff = tmp_table.timeDiff
FROM tmp_table
WHERE userId   = tmp_table.userId  
AND purchaseTime = tmp_table.purchaseTime;

谁能帮我更新 table？

Answer 1

您可以使用lag window函数来查找上次购买日期，并将两者相减即可：

SELECT userId,  
       purchaseTime,
       purchaseTime - 
        LAG(purchaseTime) OVER 
           (PARTITION BY userId ORDER BY purchaseTime) AS timeDiff
FROM   purchases

Answer 2

所以根据@mureinik 的查询，为了进行更新，您必须执行以下操作：

UPDATE purchases
SET timeDiff = tmp_table.timeDiff
FROM (SELECT userId, purchaseTime ,
       (EXTRACT(epoch FROM purchaseTime - LAG(purchaseTime) OVER 
           (PARTITION BY userId ORDER BY purchaseTime))/60)::integer AS timeDiff
        FROM   purchases) AS tmp_table
WHERE purchases.userId = tmp_table.userId
AND purchases.timeDiff = tmp_table.timeDiff;

在更新中，您将有 EXTRACT 和 epoch FROM 语句，这是为了 return 间隔中的秒数。如果你想在几分钟内将它们除以 60 </code>，最后如果你想四舍五入，只需将它转换为 <code>integer。

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