汇总列返回零

Question

我想要聚合表中的每个应用统计apps

我有以下查询，但由于某些原因，所有的结果返回0

select 
    a.id, 
    'support' as domain, 
    'summary' as type, 
    90 as interval, 
    json_build_object(
     'new', count(new), 
     'closed', count(closed_c), 
     'reply_rate', count(reply_rate), 
     'median_response', max(median_response.response_time) 
    ) as data 
from apps a 
full join (
    SELECT * from conversations c 
    WHERE c.started_at::date > (current_date - (90 || ' days')::interval)::date 
) as new on new.app_id = a.id 
full join (
    SELECT * from conversations c 
    WHERE c.closed_at::date > (current_date - (90 || ' days')::interval)::date 
) as closed_c on closed_c.app_id = a.id 
full join (
    SELECT * from conversations c 
    WHERE c.started_at::date > (current_date - (90 || ' days')::interval)::date AND c.first_response_at is not null 
) as reply_rate on reply_rate.app_id = a.id 
full join (
    SELECT c.app_id, extract(epoch from (c.first_response_at - c.started_at)) as response_time, ntile(2) OVER (ORDER BY (c.first_response_at - c.started_at)) AS bucket FROM conversations c 
    WHERE c.started_at::date > (current_date - (90 || ' days')::interval)::date AND c.first_response_at is not null 
) as median_response on median_response.app_id = a.id 
where a.test = false 
group by a.id 

Answer 1

我不能告诉到底为什么一切都是零，但

＃1：full join应left join更换（因where a.test = false）

＃2：当你与不同的条件访问同一个表四倍这可以或许通过一个单一的选择使用条件的聚集所取代。

检查这个返回正确的计数，然后离开它加入到apps。

select 
    app_id, 
    sum(new), 
    sum(closed_c), 
    sum(reply_rate), 
    max(case when bucket = 1 then response_time end) 
from 
(
    SELECT app_id, 

     1 as new, 

     case when c.closed_at::date > (current_date - (90 || ' days')::interval)::date then 1 else 0 end as closed_c, 

     case when c.first_response_at is not null then 1 else 0 end as reply_rate, 

     extract(epoch from (c.first_response_at - c.started_at)) as response_time, 

     ntile(2) OVER (ORDER BY (c.first_response_at - c.started_at)) AS bucket 
    FROM conversations c 
    -- assuming that closed_at is always after started_at 
    WHERE c.started_at::date > (current_date - (90 || ' days')::interval)::date 
) as dt 
group by app_id 

Answer 2

你为什么这样做当你不使用这些表中的单个列时，有很多联接吗？

如果你只需要来算，你可以做这样的：

select 
    a.id, 
    'support' as domain, 
    'summary' as type, 
    90 as interval, 
    json_build_object(
     'new',    (SELECT count(*)              from conversations c WHERE c.app_id = a.id and c.started_at::date > current_date - 90), 
     'closed',   (SELECT count(*)              from conversations c WHERE c.app_id = a.id and c.closed_at::date > current_date - 90), 
     'reply_rate',  (SELECT count(*)              from conversations c WHERE c.app_id = a.id and c.started_at::date > current_date - 90 and c.first_response_at is not null), 
     'median_response', (SELECT max(extract(epoch from (c.first_response_at - c.started_at))) from conversations c WHERE c.app_id = a.id and c.started_at::date > current_date - 90 and c.first_response_at is not null) 
    ) as data 
from apps a 
where a.test = false 

另外，为什么您使用的间隔加/减天到date类型？你可以做current_date - 90。

，我建议你在conversations创建一些指标也：

create index i_conversations_started_at on conversations (id, started_at::date); 
create index i_conversations_closed_at on conversations (id, closed_at::date);