sql - postgresql 中的第一个和最后一个值聚合函数可以正确处理 NULL 值

Question

我知道有聚合函数可以获取 postgresql 中行的最后一个和第一个值

我的问题是，它们无法按我的需要工作。我可以使用一个 postgresql 向导的帮助。我正在使用 postgresql 9.2 - 以防该版本更容易提供解决方案。

查询

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, TRUE, something, 2010-08-15 21:16:40, 2012-08-25 07:44:30
1, TRUE, something, 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, 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
) as bar 
group by id

然后我得到

id, active, install_date, remove_date
1, TRUE, 2008-08-02 11:13:39, 2012-08-25 07:44:30

不是

id, active, install_date, remove_date
1, TRUE, 2008-08-02 11:13:39, NULL

如我所料

如果最后一行的值为空，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？

编辑1

Roman Pekar提供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。但是我运行的查询更复杂。我意识到我的问题可能有替代解决方案 - 这就是为什么要更新帖子以包含原始的、更复杂的查询。即:

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.id, v.active
) as foo group by partner_id

如您所见，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，最后汇总这些车辆的车主的数量。

/EDIT1

Answer 1

您可以使用窗口函数 lead() and lag()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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