Oracle：如何减去两个日期并获得结果的秒数

Question

我创建了 oracle NVL function，它将减去开始和结束时间列并给出结果

但是我没有得到正确的输出值

例子


Start time- 2:33:10 

End time -2:33:20 

Actual Output - 9 seconds

Expected Output- 10 seconds

查询

NVL(TO_CHAR( table.time1,'YYYY/MM/DD HH24:MI:SS'),' ')"Time1",
NVL(TO_CHAR( table.time2,'YYYY/MM/DD HH24:MI:SS'),' ')"Time2",
NVL(dbms_lob.substr(oprm.message,4000),' ') AS "Messages",
NVL(REGEXP_SUBSTR (CAST( table.time2 AS TIMESTAMP) - CAST( table.time1 AS TIMESTAMP), '\d{2}:\d{2}:\d{2}'),' ') AS duration

Answer 1

让我看看是否可以帮助阐明您的情况。首先让我们创建一个 table

-- format the date in my current session.--
alter session set nls_date_format = 'DD-MON-YYYY HH12:MI:SS PM';

 CREATE TABLE TIME_HOLDER 
    (
      CREATED_DATE DATE 
    , SECOND_DATE DATE 
    )

向其中添加一些记录 table 以开始计算差异。

SET DEFINE OFF;
Insert into TIME_HOLDER (CREATED_DATE,SECOND_DATE) values (to_date('02-NOV-2021 02:01:45 PM','DD-MON-YYYY HH12:MI:SS PM'),to_date('22-NOV-2021 02:01:52 PM','DD-MON-YYYY HH12:MI:SS PM'));

现在要开始使用两个日期列计算两个日期之间的差异，您必须使用类似以下内容的方法。

select     extract (day from numtodsinterval (second_date
                                                   - add_months (created_date,
                                                                 floor (months_between (second_date,created_date))),
                                                   'day'))
             || ' days - '
             || extract (hour from numtodsinterval (second_date
                                                    - add_months (created_date,
                                                                  floor (months_between (second_date,created_date))),
                                                    'day'))
             || ':'
             || extract (minute from numtodsinterval (second_date
                                                      - add_months (created_date,
                                                                    floor (months_between (second_date, created_date))),
                                                      'day'))
             || ':'
             || extract (second from numtodsinterval (second_date
                                                      - add_months (created_date,
                                                                    floor (months_between (second_date, created_date))),
                                                      'day')) as time_diff
     from   TIME_HOLDER

首先您可以计算这两个日期之间的间隔，然后从该间隔导出您需要的所有数据：希望这个例子有助于阐明如何在 oracle 中做这样的事情。

Answer 2

您的数据类型很可能是 TIMESTAMP，这可以解释舍入问题。

您可以 解决方法 它首先 casting 到 DATE （以摆脱毫秒）然后将其转换返回 TIMESTAMP（以便能够执行您的 regexp_substr）

此示例数据重现了您的问题

select opa.*, 
NVL(REGEXP_SUBSTR (CAST(opa.end_time AS TIMESTAMP) - CAST(opa.start_time AS TIMESTAMP), '\d{2}:\d{2}:\d{2}'),' ') AS duration,
NVL(REGEXP_SUBSTR (CAST(CAST(opa.end_time AS DATE)AS TIMESTAMP) - CAST(CAST(opa.start_time AS DATE)AS TIMESTAMP), '\d{2}:\d{2}:\d{2}'),' ') AS duration2
from tab opa;

START_TIME                           END_TIME                             DURATION                    DURATION2                   
------------------------------------ ------------------------------------ --------------------------- ---------------------------
04.05.2021 09:13:07,555000000 +02:00 04.05.2021 09:13:18,111000000 +02:00 00:00:10                    00:00:11

Oracle：如何减去两个日期并获得结果的秒数

Oracle : how to subtract two dates and get seconds of the result

oracle

datetime

date

ceil