sql - 计算sql查询中的持续时间总和

Question

我有一个表，其中有两列开始时间和结束时间。我能够计算每行的持续时间，但我也想获得总持续时间。如何做到这一点。

谢谢

Answer 1

您的列的数据类型为 TIMESTAMP，如下所示:

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.

用一个时间戳减去另一个时间戳，得到 INTERVAL 数据类型:

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 INTERVAL 数据类型无法求和。这是一个烦人的限制:

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

要规避此限制，您可以使用秒数进行转换和计算，如下所示:

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.

然后它们就是可以求和的正常数字

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.

如果您愿意，您可以将此数字转换回 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.

问候，罗布。