Question

我必须在pg_stat_activity中汇总数据并将其作为json返回。 Postgres 9.3 onboard。需要的结果如下：

{
  "web": 67,
  "postgres": 2,
  "totalSessions": 69,
  "idle in transaction": 2,
  "active": 1,
  "idle": 66
}

其中，active是活动会话数，idle是空闲，total，两者之和，其余是每个用户的会话数。我们不需要按数据库对其进行分组，但原则是相同的。

不 - 我怎么能这样做？

Answer 1

我提出的问题并非如此：

mon=# with pre as (
  select  DISTINCT
    concat('"totalSessions":',sum(count(1)) over (),'') total
  , concat('"',usename,'":',sum(count(1)) over (partition by usename),'') u
  , concat('"',state,'":',sum(count(1)) over (partition by state),'') s
  from pg_stat_activity
  group by state,usename
)
  select concat('{',string_agg(j,','),'}')::json from (
    select distinct u j from pre u
    union all
    select distinct total from pre s
    union all
    select distinct s from pre s
  ) pg_stat_act
;
                                          concat
-------------------------------------------------------------------------------------------
 {"web":67,"postgres":2,"totalSessions":69,"idle in transaction":2,"active":1,"idle":66}
(1 row)

Time: 1.807 ms

pg_stat_activity聚合数据，用于在json

1 个答案: