how to sum jsonb values in postgres?-CodePudding

I'm working with postgres and I need to sum the value of jsonb fields

     Column      │         Type         │                             Modifiers
─────────────────┼──────────────────────┼────────────────────────────────────────────────────────────────────
 id              │ integer              │ not null default
 practice_id     │ character varying(6) │ not null
 date            │ date                 │ not null
 pct_id          │ character varying(3) │
 astro_pu_items  │ double precision     │ not null
 astro_pu_cost   │ double precision     │ not null
 total           │ jsonb                │

So I need to sum all the values for the total field for all the records and after that I need to use aggregate functions on them like min, max and avg

I started a query like this

select id, json_object_agg(key, val)
from (
    select id, key, sum(value::numeric) val
    from mytable t, jsonb_each_text(total)
    group by id, key
    ) s
group by id;

I'm confused after this. Is there a better way to solve this problem?

I need to use the aggregate function after adding all the values inside jsonb. Is that possible?

I'm getting a result like this


id   json_object_agg
25  { "a" : 1, "b" : 2 }
26  { "a" : 1, "b" : 2 }
24  { "b" : 2, "a" : 1 }

I was expecting this

 id   json_object_agg
   25    3
   26    3

Something like this

CodePudding user response：

To sum all the values of the total column for each of the rows, you need to group by id but not by key :

select t.id, sum(tt.value::numeric) val
from mytable t, jsonb_each(total) AS tt
where jsonb_typeof(tt) = 'number'  -- to avoid errors in the sum() if non numeric values
group by t.id

To calculate the min, max, avg on all the rows of the table, you don't need to group the rows :

select min(s.val), max(s.val), avg(s.val)
from
(
    select t.id, sum(tt.value::numeric) val
    from mytable t, jsonb_each(total) AS tt
    where jsonb_typeof(tt) = 'number'  -- to avoid errors in the sum() if non numeric values
    group by t.id
) as s