sql - 连接2个表计算时间差

Question

我们有几个独立事件表。 对于给定的 id，我们想知道 2 个事件之间的时间差。 现在，对于给定的 id，可以有多个相同的事件，我们只对“下一个”事件感兴趣。

例子:

http://sqlfiddle.com/#!4/1f724/6/0

架构

create table event_a (
  timestmp timestamp(3),
  id       varchar2(50)
);

create table event_b (
  timestmp timestamp(3),
  id       varchar2(50)
);

insert into event_a values (to_timestamp('2015-05-12 10:22:00', 'YYYY-MM-DD HH24:MI:SS'), 'x');
insert into event_b values (to_timestamp('2015-05-12 10:22:05', 'YYYY-MM-DD HH24:MI:SS'), 'x');
insert into event_a values (to_timestamp('2015-05-12 10:22:08', 'YYYY-MM-DD HH24:MI:SS'), 'x');
insert into event_b values (to_timestamp('2015-05-12 10:22:10', 'YYYY-MM-DD HH24:MI:SS'), 'x');

查询

这是我想出的查询，但在我看来有点太复杂了。我想知道是否有更好的方法来做到这一点。

如果这些表中有大量数据，此查询也需要很长时间才能运行。

select a.id, nvl((extract(day from (b.timestmp - a.timestmp))   * 86400
                 + extract(hour from (b.timestmp - a.timestmp))   * 3600 
                 + extract(minute from (b.timestmp - a.timestmp)) * 60   
                 + extract(second from (b.timestmp - a.timestmp)) ), 0) as duration_in_sec
from event_a a
join event_b b on a.id = b.id and b.timestmp = (
  select min(timestmp)
  from event_b
  where id = a.id
  and timestmp >= a.timestmp
)

Answer 1

这个解决方案应该更快，因为它没有自连接子查询:

select id, extract(day from (d)) * 86400 + extract(hour from (d)) * 3600 
         + extract(minute from (d)) * 60  + extract(second from (d)) duration
  from (
    select a.id, b.timestmp - a.timestmp d,
        row_number() over (partition by a.id, a.timestmp order by b.timestmp) rn
      from event_a a 
      join event_b b on a.id = b.id and a.timestmp <= b.timestmp
      order by a.timestmp)
  where rn = 1

SQLFiddle