Postgres：如何获得多列的聚合

Question

我想用postgres实现过滤导航，但我不知道如何返回结果集中多个字段中每个值的计数。

示例模式：

id, name, status 

此查询我愿意看到类似（不必模仿这种结构）：

name: [(Bob, 20), (Joe, 15), (Sue, 5)] 
status[(active, 15), (inactive, 25)] 

Answer 1

检查Grouping Sets

with t (name, status) as (values 
    ('Bob', 'active'),('Bob', 'active'), 
    ('Joe', 'inactive'),('Joe', 'active'),('Joe', 'active') 
) 
select json_object_agg(case g when 1 then 'name' else 'status' end,a) 
from (
    select jsonb_agg(jsonb_build_object(coalesce(name,status), total)) as a, g 
    from (
     select name, status, count(*) as total, grouping(name,status) as g 
     from t 
     group by grouping sets ((name),(status)) 
    ) s 
    group by g 
) s 
; 
            json_object_agg         
------------------------------------------------------------------------------------ 
{ "name" : [{"Bob": 2}, {"Joe": 3}], "status" : [{"active": 4}, {"inactive": 1}] }