Question

表DB2

ID       HOURS        HOURSMINUTES       
1000     480.5        30:30:00

我希望得到HOURS - HOURSMINUTES

ID       HOURS - HOURSMINUTES
1000       450.0

HOURS是HOURS中的浮动值 - 所以480.5小时。 HOURSMINUTES是字符串值：30：30：00（30小时30分00秒）

如何减去？

这是我的完整表达，因为我从两个表中获取值（我以这种格式获取它们但我无法减去）。我已经将HOURS作为两个时间戳格式的减法 - 结果是浮点数。 cumulativetime是字符串值。

select t1.id,
dec (( timestampdiff(
  4, 
  char(t1.actualfinish - t1.reportdate))/60.00),10,2) as HOURS,t2.cumulativetime as HOURSMINUTES
from t1
join t2 on t2.id=t1.id

当我尝试在下面插入解决方案时，我收到了错误。

    select t1.id,
    dec (( timestampdiff(
      4, 
      char(t1.actualfinish - t1.reportdate))/60.00),10,2) as HOURS,t2.cumulativetime as HOURSMINUTES,

dec (( timestampdiff(
      4, 
      char(t1.actualfinish - t1.reportdate))/60.00),10,2)- cast(substr(t2.cumulativetime, 1, 2) as int) - 
      (cast(substr(t2.cumulativetime, 4, 2) as int) / 60.0) as diff
    from t1
    join t2 on t2.id=t1.id

我也尝试了Kapil版本：

select t1.id,
    dec (( timestampdiff(
      4, 
      char(t1.actualfinish - t1.reportdate))/60.00),10,2) as HOURS,t2.cumulativetime as HOURSMINUTES,

(dec (( timestampdiff(
  4, 
  char(t1.actualfinish - t1.reportdate))/60.00),10,2) - (CAST(substr(t2.cumulativetime , 1, 2) AS float) + CAST(substr(t2.cumulativetime , 4, 2) AS float)/60 + CAST(substr(t2.cumulativetime , 7, 2) AS float)/3600)) as diff

from t1
    join t2 on t2.id=t1.id

Answer 1

select HOURS - 
       cast(substr(HOURSMINUTES, 1, 2) as int) - 
      (cast(substr(HOURSMINUTES, 4, 2) as int) / 60.0) as diff

Answer 2

试试这个：

Select  (HOURS - (CAST(substr(HOURSMINUTES , 1, 2) AS float) + CAST(substr(HOURSMINUTES , 4, 2) AS float)/60 + CAST(substr(HOURSMINUTES , 7, 2) AS float)/3600)) as diff
From table

Answer 3

以下是在兼容模式下使用DB2 for z / OS 9.1版的解决方案：

select
  t.HRS
  - cast(substr(t.HMS, 1, locate(':', t.HMS) - 1) as FLOAT)
  - (cast(substr(t.HMS, locate(':', t.HMS) + 1, locate(':', t.HMS, locate(':', t.HMS) + 1) - locate(':', t.HMS) - 1) as FLOAT) / 60.0)
  - (cast(substr(t.HMS, locate(':', t.HMS, locate(':', t.HMS, locate(':', t.HMS) + 1)) + 1) as FLOAT) / 3600.0)
from
  (
    select
      cast(480.5 as float) as HRS
      , '333:44:55' as HMS
    from
      sysibm.SYSDUMMY1
  ) as t
with ur for read only;

这会得到146.75138888888887的结果。

如果您有LOCATE_IN_STRING可用，则可以使用它来简化在:字符串中查找第n个HMS。

SQL如何减去浮点时间和字符串时间？

3 个答案: