计算SQL查询中的持续时间总和 [英] Calculate the Sum of duration in sql query
问题描述
我有一个表,该表有两列开始时间和结束时间.我能够计算每一行的持续时间,但我也想获得总持续时间.该怎么做.
I have a table which has two columns start time and end time. I am able to calculate the time duration for each row but I also want to get the total duration. how to do this.
谢谢
推荐答案
您的列的数据类型为TIMESTAMP,如下所示:
Your columns are of datatype TIMESTAMP, like this:
SQL> create table mytable (start_time,end_time)
2 as
3 select to_timestamp('2009-05-01 12:34:56','yyyy-mm-dd hh24:mi:ss')
4 , to_timestamp('2009-05-01 23:45:01','yyyy-mm-dd hh24:mi:ss')
5 from dual
6 union all
7 select to_timestamp('2009-05-01 23:45:01','yyyy-mm-dd hh24:mi:ss')
8 , to_timestamp('2009-05-02 01:23:45','yyyy-mm-dd hh24:mi:ss')
9 from dual
10 union all
11 select to_timestamp('2009-05-01 07:00:00','yyyy-mm-dd hh24:mi:ss')
12 , to_timestamp('2009-05-01 08:00:00','yyyy-mm-dd hh24:mi:ss')
13 from dual
14 /
Tabel is aangemaakt.
从另一个时间戳中减去一个时间戳,将导致INTERVAL数据类型:
Subtracting one timestamp from another, leads to an INTERVAL datatype:
SQL> select start_time
2 , end_time
3 , end_time - start_time time_difference
4 from mytable
5 /
START_TIME END_TIME TIME_DIFFERENCE
------------------------------ ------------------------------ ------------------------------
01-05-09 12:34:56,000000000 01-05-09 23:45:01,000000000 +000000000 11:10:05.000000000
01-05-09 23:45:01,000000000 02-05-09 01:23:45,000000000 +000000000 01:38:44.000000000
01-05-09 07:00:00,000000000 01-05-09 08:00:00,000000000 +000000000 01:00:00.000000000
3 rijen zijn geselecteerd.
并且不能对INTERVAL数据类型求和.这是一个令人讨厌的限制:
And INTERVAL datatypes cannot be summed. It's an annoying restriction:
SQL> select sum(end_time - start_time)
2 from mytable
3 /
select sum(end_time - start_time)
*
FOUT in regel 1:
.ORA-00932: inconsistente gegevenstypen: NUMBER verwacht, INTERVAL DAY TO SECOND gekregen
要避免这种限制,您可以使用秒数进行转换和计算,如下所示:
To circumvent this restriction, you can convert and calculate with the number of seconds, like this:
SQL> select start_time
2 , end_time
3 , trunc(end_time) - trunc(start_time) days_difference
4 , to_number(to_char(end_time,'sssss')) - to_number(to_char(start_time,'sssss')) seconds_difference
5 from mytable
6 /
START_TIME END_TIME DAYS_DIFFERENCE SECONDS_DIFFERENCE
------------------------------ ------------------------------ --------------- ------------------
01-05-09 12:34:56,000000000 01-05-09 23:45:01,000000000 0 40205
01-05-09 23:45:01,000000000 02-05-09 01:23:45,000000000 1 -80476
01-05-09 07:00:00,000000000 01-05-09 08:00:00,000000000 0 3600
3 rijen zijn geselecteerd.
然后它们是可以累加的普通数字
And then they are normal NUMBERs that can be summed
SQL> select sum
2 ( 86400 * (trunc(end_time) - trunc(start_time))
3 + to_number(to_char(end_time,'sssss')) - to_number(to_char(start_time,'sssss'))
4 ) total_time_difference
5 from mytable
6 /
TOTAL_TIME_DIFFERENCE
---------------------
49729
1 rij is geselecteerd.
如果愿意,您可以将此数字转换回INTERVAL:
And if you wish, you can convert this number back to an INTERVAL:
SQL> select numtodsinterval
2 ( sum
3 ( 86400 * (trunc(end_time) - trunc(start_time))
4 + to_number(to_char(end_time,'sssss')) - to_number(to_char(start_time,'sssss'))
5 )
6 , 'second'
7 ) time_difference
8 from mytable
9 /
TIME_DIFFERENCE
------------------------------
+000000000 13:48:49.000000000
1 rij is geselecteerd.
关于, 罗布.
这篇关于计算SQL查询中的持续时间总和的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!