Question

当减去timestamps时，返回值是interval数据类型。是否有一种优雅的方法可以将此值转换为区间中的总（毫秒/微秒），即整数。

以下可行，但不是很漂亮：

select abs( extract( second from interval_difference ) 
          + extract( minute from interval_difference ) * 60 
          + extract( hour from interval_difference ) * 60 * 60 
          + extract( day from interval_difference ) * 60 * 60 * 24
            )
  from ( select systimestamp - (systimestamp - 1) as interval_difference
           from dual )

SQL或PL / SQL中是否有更优雅的方法？

Answer 1

一种简单的方法：

select extract(day from (ts1-ts2)*86400) from dual;

我们的想法是将间隔值转换为天数86400（= 24 * 60 * 60）。然后提取'day'值，这实际上是我们想要的第二个值。

Answer 2

我希望这有帮助：

zep@dev> select interval_difference
      2        ,sysdate + (interval_difference * 86400) - sysdate as fract_sec_difference
      3  from   (select systimestamp - (systimestamp - 1) as interval_difference
      4          from   dual)
      5 ;

INTERVAL_DIFFERENCE                                                             FRACT_SEC_DIFFERENCE
------------------------------------------------------------------------------- --------------------
+000000001 00:00:00.375000                                                                 86400,375

通过测试：

zep@dev> select interval_difference
      2        ,abs(extract(second from interval_difference) +
      3        extract(minute from interval_difference) * 60 +
      4        extract(hour from interval_difference) * 60 * 60 +
      5        extract(day from interval_difference) * 60 * 60 * 24) as your_sec_difference
      6        ,sysdate + (interval_difference * 86400) - sysdate as fract_sec_difference
      7        ,round(sysdate + (interval_difference * 86400) - sysdate) as sec_difference
      8        ,round((sysdate + (interval_difference * 86400) - sysdate) * 1000) as millisec_difference
      9  from   (select systimestamp - (systimestamp - 1) as interval_difference
     10          from   dual)
     11  /

INTERVAL_DIFFERENCE                                                             YOUR_SEC_DIFFERENCE FRACT_SEC_DIFFERENCE SEC_DIFFERENCE MILLISEC_DIFFERENCE
------------------------------------------------------------------------------- ------------------- -------------------- -------------- -------------------
+000000001 00:00:00.515000                                                                86400,515            86400,515          86401            86400515

zep@dev>

Answer 3

我发现这个有用。显然，如果你使用时间戳进行算术运算，它们会转换为某种内部数据类型，当相互减去时，会将间隔作为数字返回。

容易吗？是。优雅？没有。完成工作？哦是的。

SELECT ( (A + 0) - (B + 0) ) * 24 * 60 * 60 
FROM
(
   SELECT SYSTIMESTAMP A,
          SYSTIMESTAMP - INTERVAL '1' MINUTE B
   FROM DUAL
);

Answer 4

不幸的是，我不认为有一种替代（或更优雅）的方法来计算pl / sql中间隔类型的总秒数。正如this article提到：

... unlike .NET, Oracle provides no simple equivalent to TimeSpan.TotalSeconds.

因此从区间中提取日，小时等并将它们与相应的值相乘似乎是唯一的方法。

Answer 5

基于zep's answer，为方便起见，我将内容整理成一个函数：

CREATE OR REPLACE FUNCTION intervalToSeconds( 
     pMinuend TIMESTAMP , pSubtrahend TIMESTAMP ) RETURN NUMBER IS

vDifference INTERVAL DAY TO SECOND ; 

vSeconds NUMBER ;

BEGIN 

vDifference := pMinuend - pSubtrahend ;

SELECT EXTRACT( DAY    FROM vDifference ) * 86400
     + EXTRACT( HOUR   FROM vDifference ) *  3600
     + EXTRACT( MINUTE FROM vDifference ) *    60
     + EXTRACT( SECOND FROM vDifference )
  INTO
    vSeconds 
  FROM DUAL ;

  RETURN vSeconds ;

END intervalToSeconds ;

Answer 6

使用以下查询：

select (cast(timestamp1 as date)-cast(timestamp2 as date))*24*60*60)

Answer 7

类似于@Zhaoping Lu的答案，但直接提取秒而不是从天数中获取秒。

SELECT extract(second from (end_date - start_date)) as "Seconds number"
FROM my_table

（在PostgresSQL 9.6.1上工作）

Answer 8

将时间戳转换为纳秒的更短方法。

SELECT (EXTRACT(DAY FROM (
    SYSTIMESTAMP --Replace line with desired timestamp --Maximum value: TIMESTAMP '3871-04-29 10:39:59.999999999 UTC'
- TIMESTAMP '1970-01-01 00:00:00 UTC') * 24 * 60) * 60 + EXTRACT(SECOND FROM
    SYSTIMESTAMP --Replace line with desired timestamp
)) *  1000000000 AS NANOS FROM DUAL;

NANOS
1598434427263027000

一种将纳秒转换为时间戳的方法。

SELECT TIMESTAMP '1970-01-01 00:00:00 UTC' + numtodsinterval(
    1598434427263027000 --Replace line with desired nanoseconds
/ 1000000000, 'SECOND') AS TIMESTAMP FROM dual;

TIMESTAMP
26/08/20 09:33:47,263027000 UTC

如预期的那样，上述方法的结果不受时区的影响。

一种将间隔转换为纳秒的较短方法。

SELECT (EXTRACT(DAY FROM (
    INTERVAL '+18500 09:33:47.263027' DAY(5) TO SECOND --Replace line with desired interval --Maximum value: INTERVAL '+694444 10:39:59.999999999' DAY(6) TO SECOND(9) or up to 3871 year
) * 24 * 60) * 60 + EXTRACT(SECOND FROM (
    INTERVAL '+18500 09:33:47.263027' DAY(5) TO SECOND --Replace line with desired interval
))) * 1000000000 AS NANOS FROM DUAL;

NANOS
1598434427263027000

一种将纳秒转换为间隔的方法。

SELECT numtodsinterval(
    1598434427263027000 --Replace line with desired nanoseconds
/ 1000000000, 'SECOND') AS INTERVAL FROM dual;

INTERVAL
+18500 09:33:47.263027

例如，如果要使用毫秒而不是纳秒，则将1000000000替换为1000。

我已经尝试了一些发布的方法，但是在将间隔乘以86400时遇到了一个异常“ ORA-01873：间隔的前导精度太小”，所以我决定发布一种有效的方法为我。

Answer 9

选择to_char（ENDTIME，'yyyymmddhh24missff'）-to_char（STARTTIME，'yyyymmddhh24missff'）AS DUR 从双； yyyymmddhh24miss-将提供SEC的持续时间 yyyymmddhh24mi持续时间（分钟） yyyymmddhh24-持续时间-小时 yyyymmdd天数持续时间

从间隔数据类型中提取总秒数

9 个答案: