Изравняване на агрегираните двойки ключ/стойност от JSONB поле?

Този конкретен случай

Функцията по-долу динамично създава изглед въз основа на таблица:

create or replace function create_totals_view(table_name text)
returns void language plpgsql as $$
declare
    s text;
begin
    execute format ($fmt$
        select string_agg(format('star_pu->>''%s'' "%s"', key, key), ',')
        from (
            select distinct key
            from %s, json_each(star_pu)
            order by 1
            ) s;
        $fmt$, '%s', '%s', table_name)
    into s;
    execute format('
        drop view if exists %s_view;
        create view %s_view as 
        select date, total_list_size, %s from %s', 
        table_name, table_name, s, table_name);
end $$;

Първо, създайте таблица от вашата заявка.

create table totals as

    SELECT date,
           AVG(total_list_size) AS total_list_size,
           json_object_agg(key, val) AS star_pu
    FROM (SELECT date,
                 SUM(total_list_size) AS total_list_size,
                 key, SUM(value::numeric) val FROM frontend_practicelist p,
                 jsonb_each_text(star_pu)
           GROUP BY date, key ) p
    GROUP BY date
    ORDER BY date;

След това използвайте функцията, която ще създаде изглед, наречен на таблицата с _view постфикс:

select create_totals_view('totals');

Накрая потърсете изгледа:

select * from totals_view;

Обобщено решение (за jsonb)

create or replace function create_jsonb_flat_view
    (table_name text, regular_columns text, json_column text)
    returns text language plpgsql as $$
declare
    cols text;
begin
    execute format ($ex$
        select string_agg(format('%2$s->>%%1$L "%%1$s"', key), ', ')
        from (
            select distinct key
            from %1$s, jsonb_each(%2$s)
            order by 1
            ) s;
        $ex$, table_name, json_column)
    into cols;
    execute format($ex$
        drop view if exists %1$s_view;
        create view %1$s_view as 
        select %2$s, %3$s from %1$s
        $ex$, table_name, regular_columns, cols);
    return cols;
end $$;

Употреба:

create table example (id int, name text, params jsonb);
insert into example values
(1, 'Anna', '{"height": 175, "weight": 55}'),
(2, 'Bob', '{"age": 22, "height": 188}'),
(3, 'Cindy', '{"age": 25, "weight": 48, "pretty": true}');

select create_jsonb_flat_view('example', 'id, name', 'params');

select * from example_view;

 id | name  | age | height | pretty | weight 
----+-------+-----+--------+--------+--------
  1 | Anna  |     | 175    |        | 55
  2 | Bob   | 22  | 188    |        | 
  3 | Cindy | 25  |        | true   | 48
(3 rows)