[英]Aggregated columns return zero
I want to aggregate statistics for each app in the table apps
我想汇总表
apps
程序中每个应用程序的统计信息
I have the following query, but for some reason all the results return 0
我有以下查询,但由于某种原因,所有结果均返回
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
I can't tell exactly why everything is zero, but 我无法确切地说出为什么一切都为零,但是
#1: full join
should be replaced by left join
(due to the where a.test = false
) #1:
full join
应替换为left join
(由于where a.test = false
)
#2: as you access the same table four times with different conditions this can be probably replaced by a single Select using conditional aggregation. #2:当您在不同条件下访问同一张表四次时,可以使用条件聚合将其替换为单个Select。
Check if this returns the correct counts and then Left Join it to apps
. 检查是否返回正确的计数,然后将其左加入
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
Why are you doing so many joins when you don't use not a single column from those tables? 当您不使用那些表中的单个列时,为什么要进行这么多的联接?
If you just need to count, you can do just like that: 如果您只需要计算,就可以这样做:
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
Also, why are you using interval to add/subtract days into a date
type? 另外,为什么要使用时间间隔将天数加/减为
date
类型?
You could just do current_date - 90
. 您可以只执行
current_date - 90
。
I recommend you create some indexes in conversations
too: 我建议您也在
conversations
创建一些索引:
create index i_conversations_started_at on conversations (id, started_at::date);
create index i_conversations_closed_at on conversations (id, closed_at::date);
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