Question

我有一张桌子

id  | registeredAt | organisationId |
=====================================
int | timestamptz  | int            |
-------------------------------------

对于每个组织，我希望在白天完成注册计划。

E.g。

organisationId | registrations
=================================================================
 1             | [ {"date": "...", "registrations": 10}, {... ] |
------------------------------------------------------------------
 2             | [ {"date": "...", "registrations": 12}, {... ] |
------------------------------------------------------------------

现在我可以在几个单独的查询中执行此操作（每个组织一个）使用：

select count(*) as "registrations", date_trunc('day', "registeredAt") as date
from my_table
where "organisationId" = 1
group by date;

并将其映射到每个organisationId，但我想知道在postgresql中运行这一切是否会更有效率。问题是我无法弄清楚如何正确地聚合数据以获得我想要的东西（这基本上是每个组织的注册直方图）。

PS。接下来，我不确定如何构建一个查询，这个查询会在没有注册的日子里给我零，但这是一个小问题。

Answer 1

尝试：

CREATE TYPE myrowtype AS ("date" date, registrations int);

SELECT organizationId, array_agg( row( registeredAt, cnt )::myrowtype ) as Registrations
FROM (
    select organizationId, date_trunc('day', registeredAt) as registeredAt, count(*) As cnt
    from my_table
    group by organizationId, date_trunc('day', registeredAt)
) x
GROUP BY organizationId
ORDER BY 1

如果您想以JSON格式获得结果（如问题所示），请使用：

SELECT organizationId, json_agg(  row( registeredAt, cnt )::myrowtype ) as Registrations
FROM (
    select organizationId, date_trunc('day', registeredAt) as registeredAt, count(*) As cnt
    from my_table
    group by organizationId, date_trunc('day', registeredAt)
) x
GROUP BY organizationId
ORDER BY 1

postgresql中的多维group by和数组聚合

1 个答案: