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！