在 DB2 中用午夜秒舍入

Question

也许这比我意识到的更简单，但我只是想简单地查看如下记录，然后将时间戳相加，但对于差异仅以秒为单位的分组，我的结果集是 0，但是我也想以小数形式计算秒数。

我在 AS400 iSeries 版本 7 上使用 DB2

示例数据

JOB_STATUS   |   JOB_STATUS_TIME
---------------------------------------------
 P                2019-10-02  08:47:12.362261
 P                2019-10-02  08:47:22.362261 
 P                2019-10-02  08:47:32.362261
 P                2019-10-02  08:47:42.362261
 P                2019-10-02  08:47:52.362261

查询：

SELECT
SUM
(
 CASE A1.JOB_STATUS WHEN 'P' THEN
(DAYS(A2.JOB_STATUS_TIME) - DAYS(A1.JOB_STATUS_TIME)) * 86400
 + MIDNIGHT_SECONDS(A2.JOB_STATUS_TIME) - 
 MIDNIGHT_SECONDS(A2.JOB_STATUS_TIME)
 END
 ) / 60 AS ACTIVE_MINUTES 
FROM SCHEMA.TABLE;

我尝试添加 ROUND({sum query}, 2) 但这对我不起作用，仍然返回零

Answer 1

如果除法运算中使用的所有列和值都是 INGEGER（或 BIGINT 或 SMALLINT），则除法将是整数除法。

您需要将某些内容转换为 DECIMAL 或使用 DECFLOAT 值的小数浮点数来获得小数除法。例如。尝试除以 60.0

SELECT
   SUM(CASE A1.JOB_STATUS WHEN 'P' THEN
        (DAYS(A2.JOB_STATUS_TIME) - DAYS(A1.JOB_STATUS_TIME)) * 86400
       + MIDNIGHT_SECONDS(A2.JOB_STATUS_TIME)
       - MIDNIGHT_SECONDS(A1.JOB_STATUS_TIME)
 END
 ) / 60.0 AS ACTIVE_MINUTES 
FROM
   SCHEMA.TABLE

https://www.ibm.com/support/knowledgecenter/en/ssw_ibm_i_72/db2/rbafz2intoper.htm

If both operands of an arithmetic operator are integers with zero scale, the operation is performed in binary, and the result is a large integer unless either (or both) operand is a big integer, in which case the result is a big integer. Any remainder of division is lost

Answer 2

有点跑题了，不过我觉得你做的可以用

TIMESTAMPDIFF(2, MAX(JOB_STATUS_TIME) - MIN(JOB_STATUS_TIME) )

在 DB2 中用午夜秒舍入

Rounding with midnight seconds in DB2

sql

db2

db2-400