获取日期,小时,分钟和秒的日期差异(上午/下午日期)

Question

试图从日期day - hours - mins - sec的字符串条目中获取日期差异.时间条目也有AM, PM.我正在尝试使用子字符串并将其转换为日期时间,但是AM - PM使其变得困难.

我有:06-OCT-15 09.30.20.000000 AM和06-OCT-15 09.45.20.000000 PM

我想要:0 day(s) 12 hour(s) 15 min(s) 0 sec(s).

Answer 1

请参阅下面的SQL Server查询.我知道这会有效,因为我有一些经验.

WITH CTE AS (
    SELECT CAST(
      DATEDIFF(
              DAY,
              '2015-10-06 08:00:00.00 AM',
              '2015-10-06 09:00:00.00 PM'
          ) AS 
          VARCHAR(1000)
      ) + ' day(s)' AS      Days,
      CAST(
          DATEDIFF(
              hh,
              '2015-10-06 08:00:00.00 AM',
              '2015-10-06 09:00:00.00 PM'
          ) AS VARCHAR(1000)
      ) + ' hour(s)' AS     Hours,
      CAST(
          DATEDIFF(
              mi,
              '2015-10-06 08:00:00.00 AM',
              '2015-10-06 09:00:00.00 PM'
          ) AS VARCHAR(1000)
      ) + ' min(s)' AS      Minutes,
      CAST(
          DATEDIFF(
              ss,
              '2015-10-06 08:00:00.00 AM',
              '2015-10-06 09:00:00.00 PM'
          ) AS VARCHAR(1000)
      ) + ' sec(s)' AS Seconds
)
SELECT Days + ' ' + Hours + ' ' + Minutes + ' ' + Seconds AS FullDifference
FROM   CTE