Question

我有一个包含两个DATE列END_TIME和START_TIME的表格。在此表中，我运行以下查询：

SELECT y.ID,
       TO_CHAR(  TO_DATE('00:00:00', 'HH24:MI:SS') + (y.END_TIME - y.START_TIME) 
               , 'HH24:MI:SS') AS RUNTIME,
       y.END_TIME - y.START_TIME AS RUNTIME2,
       TO_CHAR(y.START_TIME, 'DD-MON-YYYY HH24:MI:SS') AS START_TIME,
       TO_CHAR(y.END_TIME, 'DD-MON-YYYY HH24:MI:SS') AS END_TIME
FROM mytable y;

我得到这两行：

ID | RUNTIME | RUNTIME2                                   | START_TIME          | END_TIME
------------------------------------------------------------------------------------------------------
 1 | 04:26:17| 0.1849189814814814814814814814814814814815 | 30-JAN-2015 19:45:48| 31-JAN-2015 00:12:05
 2 | 03:28:18| 1.14465277777777777777777777777777777778   | 06-FEB-2015 20:47:22| 08-FEB-2015 00:15:40

如您所见，ID 2的运行时间大于24小时。如何更改我的查询，以便RUNTIME的{{1}}代替ID 2？

Answer 1

你需要将时差分成它的组成日，小时，分钟和秒元素，将天数* 24与小时数相结合，并将它重新组合在一起。

当减去日期时，你会得到差异作为天数，所以你需要将小数部分转换成其他元素，你可以用trunc和mod的组合来做;使用CTE使这更容易遵循并显示每个值自己以及合并为一个字符串：

with y as (
  select id, end_time - start_time as runtime
  from mytable
)
select id,
  runtime,
  trunc(runtime) as days,
  24 * trunc(runtime) as day_hours,
  trunc(24 * mod(runtime, 1)) as hours,
  trunc(60 * mod(24 * (runtime), 1)) as minutes,
  trunc(60 * mod(24 * 60 * (runtime), 1)) as seconds,
  lpad(24 * trunc(runtime)
    + trunc(24 * mod(runtime, 1)), 2, '0')
    ||':'|| lpad(trunc(60 * mod(24 * (runtime), 1)), 2, '0')
    ||':'|| lpad(trunc(60 * mod(24 * 60 * (runtime), 1)), 2, '0')
    as runtime
from y;

        ID    RUNTIME       DAYS  DAY_HOURS      HOURS    MINUTES    SECONDS RUNTIME 
---------- ---------- ---------- ---------- ---------- ---------- ---------- --------
         1 .184918981          0          0          4         26         16 04:26:16 
         2 1.14465278          1         24          3         28         18 27:28:18

您还可以将日期转换为计算的时间戳，这会为您提供间隔类型，然后使用extract函数来获取元素;虽然校长是相同的：

with y as (
  select id,
    cast(end_time as timestamp) - cast (start_time as timestamp) as runtime
  from mytable
)
select id,
  runtime,
  extract (day from runtime) as days,
  24 * extract (day from runtime) as day_hours,
  extract (hour from runtime) as hours,
  extract (minute from runtime) as minutes,
  extract (second from runtime) as seconds,
  lpad(24 * extract (day from runtime) + extract (hour from runtime), 2, '0')
    ||':'|| lpad(extract (minute from runtime), 2, '0')
    ||':'|| lpad(extract (second from runtime), 2, '0')
    as runtime
from y;

        ID RUNTIME           DAYS  DAY_HOURS      HOURS    MINUTES    SECONDS RUNTIME 
---------- ----------- ---------- ---------- ---------- ---------- ---------- --------
         1 0 4:26:17.0          0          0          4         26         17 04:26:17 
         2 1 3:28:18.0          1         24          3         28         18 27:28:18

或略有变化，从日期中获得差异，然后将其转换为间隔：

with y as (
  select id,
    numtodsinterval(end_time - start_time, 'DAY') as runtime
  from mytable
)
...

SQL Fiddle demo

Oracle显示超过24小时

1 个答案: