问

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

camera98 发布于 2023-02-13 11:59

post

我知道在postgresql中有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, 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

然后我明白了

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

不

id, active, install_date, remove_date
1, TRUE, 2008-08-02 11:13:39, 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为我的问题提供了替代解决方案,但这不符合我的需求.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.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

1 个回答

您可以使用窗口函数lead()并lag()检查第一条和最后一条记录，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

sql fiddle demo

2023-02-13 12:01 回答

酱油丸子-310

撰写答案

今天，你开发时遇到什么问题呢？

立即提问

热门标签