如何获取postgreSQL中两个值之间的总计数

Question

我在一个列中有 3 个活动 (completed_order, email, viewed)，我想计算每个 completed_order activity 之间发生了多少 email 个活动，并将它们保存在不同的专栏。我写了这个查询：

SELECT  activity_id, ts, customer, activity ,
        case when activity = 'completed_order'      
        then count(*)filter (where activity = 'email' ) over (partition by customer order by ts )
        else null end as Aggregate_in_between
    FROM public.activity_stream as az1  where customer = 'Lehmanns Marktstand' order by ts ;

我通过上述查询得到以下结果。

activity_id	ts	customer	activity	agg_in_btw
11089	"1996-08-12 00:00:00+05"	"Lehmanns Marktstand"	"completed_order"	0
10279	"1996-08-13 00:00:00+05"	"Lehmanns Marktstand"	"completed_order"	0
11077	"1996-08-14 00:00:00+05"	"Lehmanns Marktstand"	"email"
11092	"1996-08-17 00:00:00+05"	"Lehmanns Marktstand"	"viewed_page"
11088	"1996-08-18 00:00:00+05"	"Lehmanns Marktstand"	"viewed_page"
10284	"1996-08-19 00:00:00+05"	"Lehmanns Marktstand"	"completed_order"	1
11078	"1996-08-20 00:00:00+05"	"Lehmanns Marktstand"	"email"
11079	"1996-08-21 00:00:00+05"	"Lehmanns Marktstand"	"email"
11080	"1996-10-21 00:00:00+05"	"Lehmanns Marktstand"	"email"
10343	"1996-10-31 00:00:00+05"	"Lehmanns Marktstand"	"completed_order"	4
11090	"1996-11-01 00:00:00+05"	"Lehmanns Marktstand"	"viewed_page"
11091	"1996-11-02 00:00:00+05"	"Lehmanns Marktstand"	"email"
10497	"1997-04-04 00:00:00+05"	"Lehmanns Marktstand"	"completed_order"	5
10522	"1997-04-30 00:00:00+05"	"Lehmanns Marktstand"	"completed_order"	5

我想要的结果应该是这样的

activity_id	ts	customer	activity	agg_in_btw
11089	"1996-08-12 00:00:00+05"	"Lehmanns Marktstand"	"completed_order"	0
10279	"1996-08-13 00:00:00+05"	"Lehmanns Marktstand"	"completed_order"	1
11077	"1996-08-14 00:00:00+05"	"Lehmanns Marktstand"	"email"
11092	"1996-08-17 00:00:00+05"	"Lehmanns Marktstand"	"viewed_page"
11088	"1996-08-18 00:00:00+05"	"Lehmanns Marktstand"	"viewed_page"
10284	"1996-08-19 00:00:00+05"	"Lehmanns Marktstand"	"completed_order"	3
11078	"1996-08-20 00:00:00+05"	"Lehmanns Marktstand"	"email"
11079	"1996-08-21 00:00:00+05"	"Lehmanns Marktstand"	"email"
11080	"1996-10-21 00:00:00+05"	"Lehmanns Marktstand"	"email"
10343	"1996-10-31 00:00:00+05"	"Lehmanns Marktstand"	"completed_order"	1
11090	"1996-11-01 00:00:00+05"	"Lehmanns Marktstand"	"viewed_page"
11091	"1996-11-02 00:00:00+05"	"Lehmanns Marktstand"	"email"
10497	"1997-04-04 00:00:00+05"	"Lehmanns Marktstand"	"completed_order"	0
10522	"1997-04-30 00:00:00+05"	"Lehmanns Marktstand"	"completed_order"	0

Answer 1

试试这个方法：

select  activity_id, ts, customer, activity ,
case when activity = 'completed_order' then
sum(case when activity='email' then 1 else 0 end) over (partition by customer,grp1)
else
null
end "Aggregate_in_between"
from (
select *, sum(grp) over (partition by customer order by ts) "grp1" from (
select *, case when activity='completed_order' then 1 else 0 end "grp" from activity_stream order by ts  
) t ) q

在这里你必须为每 completed_order 次出现创建一个组，他们计算每个组中的 email 次。

您可以根据自己的方便添加 where 子句

DEMO

Answer 2

这是一个分组问题。根据“已完成”的累积计数分配一个组。然后在每组内数数：

select a.*,
       count(*) filter (where activity = 'email') over (partition by co_grp)
from (select a.*,
             count(*) filter (where activity = 'completed_order') over (partition by customer order by ts) as co_grp
      from public.activity_stream a
      where customer = 'Lehmanns Marktstand'
     ) a
order by ts ;

在您的示例数据中，您似乎只希望在已完成的订单行上执行此操作，因此为此使用 case 表达式：

select a.*,
       (case when activity = 'completed_order'
             then count(*) filter (where activity = 'email') over (partition by co_grp)
        end) as agg_in_btw
from (select a.*,
             count(*) filter (where activity = 'completed_order') over (partition by customer order by ts) as co_grp
      from public.activity_stream a
      where customer = 'Lehmanns Marktstand'
     ) a
order by ts ;

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