Search and update a JSON array element in Postgres - postgresql

Question

I have a Jsonb column that store array of elements like the following:
[
{"id": "11", "name": "John", "age":"25", ..........},
{"id": "22", "name": "Mike", "age":"35", ..........},
{"id": "33", "name": "Tom", "age":"45", ..........},
.....
]
I want to replace the 2nd object(id=22) with a totally new object. I don't want to update each property one by one because there are many properties and their values all could have changed. I want to just identify the 2nd element and replace the whole object.
I know there is a jsonb_set(). However, to update the 2nd element, I need to know its array index=1 so I can do the following:
jsonb_set(data, '{1}', '{"id": "22", "name": "Don", "age":"55"}',true)
But I couldn't find any way to search and get that index. Can someone help me out?

Answer 1

One way I can think of is to combine row_number and json_array_elements:
-- test data
create table test (id integer, data jsonb);
insert into test values (1, '[{"id": "22", "name": "Don", "age":"55"}, {"id": "23", "name": "Don2", "age":"55"},{"id": "24", "name": "Don3", "age":"55"}]');
insert into test values (2, '[{"id": "32", "name": "Don", "age":"55"}, {"id": "33", "name": "Don2", "age":"55"},{"id": "34", "name": "Don3", "age":"55"}]');
select subrow, id, row_number() over (partition by id)
from (
select json_array_elements(data) as subrow, id
from test
) as t;
subrow | id | row_number
------------------------------------------+----+------------
{"id": "22", "name": "Don", "age":"55"} | 1 | 1
{"id": "23", "name": "Don2", "age":"55"} | 1 | 2
{"id": "24", "name": "Don3", "age":"55"} | 1 | 3
{"id": "32", "name": "Don", "age":"55"} | 2 | 1
{"id": "33", "name": "Don2", "age":"55"} | 2 | 2
{"id": "34", "name": "Don3", "age":"55"} | 2 | 3
-- apparently you can filter what you want from here
select subrow, id, row_number() over (partition by id)
from (
select json_array_elements(data) as subrow, id
from test
) as t
where subrow->>'id' = '23';
In addition, think about your schema design. It may not be the best idea to store your data this way.