Teradata 中分区 window 的时差(以小时和秒为单位)(会话记录)

Time difference in hours and seconds over a partition window in Teradata (Sessionizing Records)

给出这样的 table:

cust_id                time
    123 2015-01-01 12:15:05
    123 2015-01-01 12:17:06
    123 2015-01-02 13:15:08
    123 2015-01-02 15:15:10
    456 2015-01-01 10:15:05
    456 2015-01-01 12:15:07
    456 2015-01-01 14:11:10

我想通过 cust_id 计算每个前面的记录(想想 lag 函数)之间的时间差。我想要的输出:

cust_id                time diff_hours  diff_seconds
    123 2015-01-01 12:15:05       NULL          NULL
    123 2015-01-01 12:17:06       0.00           121
    123 2015-01-02 13:15:08       1.04         89882
    123 2015-01-02 15:15:10       0.08          7202
    456 2015-01-01 10:15:05       NULL          NULL
    456 2015-01-01 12:15:07       0.08          7202 
    456 2015-01-01 14:11:10       0.08          6963

如何在 Teradata 中完成此操作?

我尝试过类似的东西:

SELECT
  *
  , (time - time) OVER (PARTITION BY cust_id ORDER BY time ROWS BETWEEN 1 PRECEDING AND CURRENT ROW)
FROM
  table_01

然而,虽然 NULL 出现在预期的位置,但我一直收到所有其他结果的 0.0。我也尝试过将 (time - time)SUM 一起使用,我也尝试过使用 EXTRACT(SECOND FROM TIME) 和其他一些变体——例如,尝试放置 DAY(4) to SECOND,但我不能似乎 syntax/ordering/conversion 非常正确,尤其是将 window 函数混入其中时。

Teradata 中没有LAG,但您可以重写它:

SELECT
  t.*
  , (time)
    - min(time) 
      OVER (PARTITION BY cust_id 
            ORDER BY time
            ROWS BETWEEN 1 PRECEDING AND 1 PRECEDING) SECOND(4)
FROM
  table_01 as t

当您尝试获取秒数时,您会遇到 "Interval Overflow" 错误,即超过 9999 秒。要么更改为 DAY(4) TO SECOND,要么使用我几年前写的这个 SQL UDF,用于计算两个时间戳的秒数之差:

REPLACE FUNCTION TimeStamp_Diff_Seconds
(
   ts1 TIMESTAMP(6)
  ,ts2 TIMESTAMP(6)
)
RETURNS DECIMAL(18,6)
LANGUAGE SQL
CONTAINS SQL
RETURNS NULL ON NULL INPUT
DETERMINISTIC
SQL SECURITY DEFINER
COLLATION INVOKER
INLINE TYPE 1
RETURN
(CAST((CAST(ts2 AS DATE)- CAST(ts1 AS DATE)) AS DECIMAL(18,6)) * 60*60*24)
      + ((EXTRACT(  HOUR FROM ts2) - EXTRACT(  HOUR FROM ts1)) * 60*60)
      + ((EXTRACT(MINUTE FROM ts2) - EXTRACT(MINUTE FROM ts1)) * 60)
      +  (EXTRACT(SECOND FROM ts2) - EXTRACT(SECOND FROM ts1))
;