postgresql中的第一个和最后一个值聚合函数,它们可以正确地使用NULL值

我知道在 postgresql中有用于获取行的最后和第一个值的聚合函数

我的问题是,他们不能按我的需要工作.我可以使用一个postgresql向导的帮助.我正在使用postgresql 9.2 – 以防该版本提供解决方案easyyer.

询问

select v.id,v.active,v.reg_no,p.install_date,p.remove_date 
from vehicle v 
    left join period p on (v.id = p.car_id) 
where v.id = 1 
order by v.id,p.install_date asc

返回6行：

id,active,reg_no,install_date,remove_date
1,TRUE,something,2008-08-02 11:13:39,2009-02-09 10:32:32
....
1,2010-08-15 21:16:40,2012-08-25 07:44:30
1,2012-09-10 17:05:12,NULL

但是当我使用聚合查询时：

select max(id) as id,last(active) as active,first(install_date) as install_date,last(remove_date) as remove_date 
from (
    select v.id,p.remove_date 
    from vehicle v 
      left join period p on (v.id = p.car_id) 
    where v.id = 1 
    order by v.id,p.install_date asc
) as bar 
group by id

然后我明白了

id,2012-08-25 07:44:30

不

id,NULL

正如我所料

如果最后一行的值为null,而不是最后一个现有值,是否可以以某种方式更改聚合函数以产生NULL？

EDIT1

Roman Pekar提出alternative solution我的问题,但这不符合我的需要.原因是 – 我简化了原始查询.但我运行的查询更复杂.我意识到可能有我的问题的替代解决方案 – 这就是为什么更新帖子以包括原始的,更复杂的查询.这是：

select partner_id,sum(active) as active,sum(installed) as installed,sum(removed) as removed 
from (
    select 
    pc.partner_id as partner_id,v.id,CASE WHEN v.active = TRUE THEN 1 ELSE 0 END as active,CASE WHEN first(p.install_date) BETWEEN '2013-12-01' AND '2014-01-01' THEN 1 ELSE 0 END as installed,CASE WHEN last(p.remove_date) BETWEEN '2013-12-01' AND '2014-01-01' THEN 1 ELSE 0 END as removed 
    from vehicle v 
        left join period p on (v.id = p.car_id) 
        left join partner_clients pc on (pc.account_id = v.client_id) 
    group by pc.partner_id,v.active
) as foo group by partner_id

正如您所看到的,我实际上需要获得几辆车的第一个和最后一个值而不是一个,并最终汇总这些车辆的所有者的车辆数量.

/ EDIT1

解决方法

您可以使用窗口函数 lead() and lag()来检查第一个和最后一个记录,例如：

select
    max(a.id) as id,max(a.first) as first,max(a.last) as last
from (
    select
         v.id,case when lag(v.id) over(order by v.id,p.install_date) is null then p.install_date end as first,case when lead(v.id) over(order by v.id,p.install_date) is null then p.remove_date end as last
    from vehicle v 
       left join period p on (v.id = p.car_id) 
    where v.id = 1 
) as a

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