Question

 event            start                             end
 event147         22-JUL-18 11.55.02.000000000  23-JUL-18 12.45.12.000000000 
 event148         24-JUL-18 13.12.45.000000000  25-JUL-18 18.33.05.000000000

上面是md_events表，我需要在事件结束147和事件148开始之间得到天数（没有时间部分）。两者都是时间戳。

我试过，但是收到错误（与numtodsinterval相同）

select dsintervaltonum(me2.start- me1.end, 'day') as days
  from md_events me1, md_events me2
 where me1.event  = 'event147'
   and me2.event =  'event148'

此查询还返回时间部分

 select to_char(me2.start- me1.end, 'DDD') as days
      from md_events me1, md_events me2
     where me1.event  = 'event147'
       and me2.event =  'event148'

Answer 1

将两个时间戳转换为日期，然后直接区分它们：

call :LIB_TIME

echo Date is: %_DATE%
echo Time is: %_TIME_SC%
echo.
echo Date is: %_YYYY%-%_MM%-%_DD%
echo Time is: %_HR%:%_MN%:%_SC%

Answer 2

你也可以使用这个：

SELECT EXTRACT(DAY FROM (me2.start- me1.end))
FROM md_events me1
    CROSS JOIN md_events me2
where
    me1.event  = 'event147' and
    me2.event =  'event148'

Answer 3

我的系统中没有Oracle。所以我将提供一种方法/逻辑来使用SQLITE找到您的解决方案。使用ORACLE语法改变SQLITE语法!!

SELECT y.start,x.endd, y.start-x.endd as diff FROM (SELECT a.* FROM NAMES AS a WHERE a.event='event147') AS x, (SELECT b.* FROM NAMES AS b WHERE b.event='event148') AS y;

Oracle SQL在两个时间戳之间的天数差异

3 个答案: