匯總列返回零

Question

我想匯總表apps程序中每個應用程序的統計信息

我有以下查詢，但由於某種原因，所有結果均返回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

我無法確切地說出為什么一切都為零，但是

＃1： full join應替換為left join （由於where a.test = false ）

＃2：當您在不同條件下訪問同一張表四次時，可以使用條件聚合將其替換為單個Select。

檢查是否返回正確的計數，然后將其左加入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

當您不使用那些表中的單個列時，為什么要進行這么多的聯接？

如果您只需要計算，就可以這樣做：

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，為什么要使用時間間隔將天數加/減為date類型？

您可以只執行current_date - 90 。

我建議您也在conversations創建一些索引：

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