彙總列返回零

Question

我想要聚合表中的每個應用統計apps

我有以下查詢，但由於某些原因，所有的結果返回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

我不能告訴到底爲什麼一切都是零，但

＃1：full join應left join更換（因where a.test = false）

＃2：當你與不同的條件訪問同一個表四倍這可以或許通過一個單一的選擇使用條件的聚集所取代。

檢查這個返回正確的計數，然後離開它加入到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

你爲什麼這樣做當你不使用這些表中的單個列時，有很多聯接嗎？

如果你只需要來算，你可以做這樣的：

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，爲什麼您使用的間隔加/減天到date類型？你可以做current_date - 90。

，我建議你在conversations創建一些指標也：

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