我正在尝试减去两个日期并期望返回一些浮动值。但我得到的回报如下:
+000000000 00:00:07.225000
将该值乘以 86400(我想以秒为单位)得到更奇怪的返回值:
+000000007 05:24:00.000000000
任何的想法?我怀疑这与类型转换有关。
我正在尝试减去两个日期并期望返回一些浮动值。但我得到的回报如下:
+000000000 00:00:07.225000
将该值乘以 86400(我想以秒为单位)得到更奇怪的返回值:
+000000007 05:24:00.000000000
任何的想法?我怀疑这与类型转换有关。
我猜你的列被定义为timestamp
而不是date
.
减去时间戳的结果是一个interval
,而减去date
列的结果是一个数字,表示两个日期之间的天数。
这记录在手册中: http:
//docs.oracle.com/cd/E11882_01/server.112/e41084/sql_elements001.htm#i48042
因此,当您将时间戳列转换为日期时,您应该得到您期望的结果:
with dates as (
select timestamp '2012-04-27 09:00:00' as col1,
timestamp '2012-04-26 17:35:00' as col2
from dual
)
select col1 - col2 as ts_difference,
cast(col1 as date) - cast(col2 as date) as dt_difference
from dates;
编辑:
如果您想转换间隔,例如秒数(作为数字),您可以执行以下操作:
with dates as (
select timestamp '2012-04-27 09:00:00.1234' as col1,
timestamp '2012-04-26 17:35:00.5432' as col2
from dual
)
select col1 - col2 as ts_difference,
extract(hour from (col1 - col2)) * 3600 +
extract(minute from (col1 - col2)) * 60 +
(extract(second from (col1 - col2)) * 1000) / 1000 as seconds
from dates;
上面的结果是55499.5802
阅读这篇文章: http ://asktom.oracle.com/pls/asktom/ASKTOM.download_file?p_file=6551242712657900129
例子:
create table yourtable(
date1 date,
date2 date
)
SQL> select datediff( 'ss', date1, date2 ) seconds from yourtable
SECONDS
----------
6269539