在sql查询中计算持续时间的总和

Question

我有一个包含两列开始时间和结束时间的表格。我能够计算每一行的持续时间，但我也想获得总持续时间。这个怎么做。

感谢

Answer 1

你列的数据类型TIMESTAMP，像这样：

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. 

减去另一个时间戳，导致间隔数据类型：

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. 

和间隔时间数据类型不能相加。这是一个恼人的限制：

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 

为了规避这个限制，你可以转换和计算与秒数，这样的：

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，它们是可以概括正规数

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. 

如果你愿意，你可以这个数字转换回间隔：

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. 

Regards， Rob。

Answer 2

你可以（至少它的工作原理上的Oracle DB）使用此查询：

select sum(end_date - start_date) from your_table 

Answer 3

编辑：前加入TRUNC（）求和基于罗布面包车Wijk出色的答复。

要找到每行时间：

select 
    end_date-start_date as DurationDays, 
    (end_date-start_date)*24 as DurationHours, 
    (end_date-start_date)*24*60 as DurationMinutes, 
    (end_date-start_date)*24*60*60 as DurationSeconds 
from your_table 

要查找的总持续时间：

select 
    sum(trunc(end_date-start_date)) as TotalDurationDays 
from your_table 

要在一个查询一举两得：

select 
    end_date-start_date as DurationDays, 
    (select sum(trunc(end_date-start_date)) from your_table) as TotalDurationDays 
from your_table 

Answer 4

这对Oracle方法简单，一个黑客攻击的一位，但：

select sum((end_timestamp+0) - (start_timestamp+0)) 
from your_table 

结果是天的NUMBER（有小数部分的小时，分​​钟和你知道的）。

我不知道什么时间戳+ 0确实;也许ANSI时间戳被转换为Oracle的早期时间戳类型，允许简单的算术。