SQL滞后或潜在客户 - 如何根据具有日期/时间差异的行选择记录

Question

我有一个包含数百万条记录的表，其中包含日期/时间戳 - 我试图返回具有差异的记录>给定的秒数。

表中包含日期/时间戳，但我无法使用where和计算的差异。 （createddate是日期字段，updatedon是日期/时间）

SELECT createddate, 
     LEAD(createddate, 1) OVER (ORDER BY createddate) AS created_next, 
     LEAD(createddate, 1) OVER (ORDER BY createddate) - createddate AS created_diff, 
     LEAD(updatedon, 1) OVER (ORDER BY updatedon) AS created_next, 
     LEAD(createddate, 1) OVER (ORDER BY updatedon) - updatedon AS Updatedon_diff 
FROM gsdaudit 
--WHERE created_diff > 1000 
ORDER BY updatedon_diff 

Answer 1

最简单的方法（假设SQL服务器）可能是使用DATEDIFF获得在几秒钟之差，并把计算的公共表表达式内;

WITH cte AS (
    SELECT 
    createddate, 
    DATEDIFF(second, createddate, 
      LEAD(createddate, 1) OVER (ORDER BY createddate)) created_diff, 
    updatedon, 
    DATEDIFF(second, createddate, 
      LEAD(updatedon, 1) OVER (ORDER BY updatedon )) updatedon_diff 
    FROM gsdaudit 
) 
SELECT * FROM cte 
WHERE created_diff > 10 OR updatedon_diff > 10 

An SQLfiddle to test with。

编辑：相同的查询为Oracle;

WITH cte AS (
    SELECT 
    createddate, 
    (LEAD(createddate, 1) OVER (ORDER BY createddate)-createddate)*24*60*60 created_diff, 
    updatedon, 
    (LEAD(updatedon, 1) OVER (ORDER BY updatedon)-updatedon)*24*60*60 updatedon_diff 
    FROM gsdaudit 
) 
SELECT * FROM cte 
WHERE created_diff > 10 OR updatedon_diff > 10 

Another SQLfiddle。