SQL 从时间戳差异中提取值

Question

我正在尝试从减去 Oracle 数据库中的两个时间戳的输出中提取天、小时、分钟、秒。然后我想获取提取的值并将它们放入单独的列中。我可以使用 substr 完成此操作，但这似乎效率不高。是否有更有效和程序化的方式来提取值？以下是具有当前和所需输出的示例查询。

示例:

SELECT 

to_timestamp('2019-11-10 15:00:00', 'YYYY-MM-DD hh24:mi:ss') -
to_timestamp('2019-10-25 13:25:00', 'YYYY-MM-DD hh24:mi:ss')  
as TIME_DIFF,

SUBSTR(to_timestamp('2019-11-10 15:00:00', 'YYYY-MM-DD hh24:mi:ss') -
to_timestamp('2019-10-25 13:25:00', 'YYYY-MM-DD hh24:mi:ss'), 9, 2)  
as DAYS

from dual

当前输出:

TIME_DIFF                     | DAYS
------------------------------+-----
+000000016 01:35:00.000000000 | 16

期望的输出:

DAYS | HOUR | MIN | SS
-----+------+-----+---+
16   |  01  |  35 | 00

Answer 1

您可以使用 extract() 从区间中提取所需的值:

with t as (
    select to_timestamp('2019-11-10 15:00:00', 'YYYY-MM-DD hh24:mi:ss') -
        to_timestamp('2019-10-25 13:25:00', 'YYYY-MM-DD hh24:mi:ss')  
        as TIME_DIFF
    from dual
)
select 
    extract(day from time_diff) days,
    extract(hour from time_diff) hours,
    extract(minute from time_diff) minutes,
    extract(second from time_diff) seconds
from t

Demo on DB Fiddle :

DAYS | HOURS | MINUTES | SECONDS---: | ----: | ------: | ------:  16 |     1 |      35 |       0