Check if a Postgres JSON array contains a string
As of PostgreSQL 9.4, you can use the ?
operator:
select info->>'name' from rabbits where (info->'food')::jsonb ? 'carrots';
You can even index the ?
query on the "food"
key if you switch to the jsonb type instead:
alter table rabbits alter info type jsonb using info::jsonb;
create index on rabbits using gin ((info->'food'));
select info->>'name' from rabbits where info->'food' ? 'carrots';
Of course, you probably don't have time for that as a full-time rabbit keeper.
Update: Here's a demonstration of the performance improvements on a table of 1,000,000 rabbits where each rabbit likes two foods and 10% of them like carrots:
d=# -- Postgres 9.3 solution
d=# explain analyze select info->>'name' from rabbits where exists (
d(# select 1 from json_array_elements(info->'food') as food
d(# where food::text = '"carrots"'
d(# );
Execution time: 3084.927 ms
d=# -- Postgres 9.4+ solution
d=# explain analyze select info->'name' from rabbits where (info->'food')::jsonb ? 'carrots';
Execution time: 1255.501 ms
d=# alter table rabbits alter info type jsonb using info::jsonb;
d=# explain analyze select info->'name' from rabbits where info->'food' ? 'carrots';
Execution time: 465.919 ms
d=# create index on rabbits using gin ((info->'food'));
d=# explain analyze select info->'name' from rabbits where info->'food' ? 'carrots';
Execution time: 256.478 ms
You could use @> operator to do this something like
SELECT info->>'name'
FROM rabbits
WHERE info->'food' @> '"carrots"';
Not smarter but simpler:
select info->>'name' from rabbits WHERE info->>'food' LIKE '%"carrots"%';