Snowball
Snowball

Reputation: 11696

Check if a Postgres JSON array contains a string

I have a table to store information about my rabbits. It looks like this:

create table rabbits (rabbit_id bigserial primary key, info json not null);
insert into rabbits (info) values
  ('{"name":"Henry", "food":["lettuce","carrots"]}'),
  ('{"name":"Herald","food":["carrots","zucchini"]}'),
  ('{"name":"Helen", "food":["lettuce","cheese"]}');

How should I find the rabbits who like carrots? I came up with this:

select info->>'name' from rabbits where exists (
  select 1 from json_array_elements(info->'food') as food
  where food::text = '"carrots"'
);

I don't like that query. It's a mess.

As a full-time rabbit-keeper, I don't have time to change my database schema. I just want to properly feed my rabbits. Is there a more readable way to do that query?

Upvotes: 272

Views: 319902

Answers (10)

Hadi Salehy
Hadi Salehy

Reputation: 360

You can use the POSITION() function or the LIKE operator with wildcard characters.

SELECT * FROM your_table WHERE (POSITION( "Feild1" IN "Feild2") > 0) ;

or

SELECT * FROM your_table WHERE "Feild2" like  '%' ||  "Feild1" || '%';

Upvotes: 0

bmusonda
bmusonda

Reputation: 701

In order to select the specific key in the JSONB, you should use ->.

select * from rabbits where (info->'food')::jsonb ? 'carrots';

Upvotes: 8

martinkaburu
martinkaburu

Reputation: 586

You can do a direct type cast from jsonb to text incase you want to check the full json and not one key.

select * from table_name
where 
column_name::text ilike '%Something%';

Upvotes: 13

M. Hamza Rajput
M. Hamza Rajput

Reputation: 10296

This might help.

SELECT a.crops ->> 'contentFile' as contentFile
FROM ( SELECT json_array_elements('[
    {
        "cropId": 23,
        "contentFile": "/menu/wheat"
    },
    {
        "cropId": 25,
        "contentFile": "/menu/rice"
    }
]') as crops ) a
WHERE a.crops ->> 'cropId' = '23';

OUTPUT:

/menu/wheat

Upvotes: 1

lmb
lmb

Reputation: 341

If the array is at the root of the jsonb column, i.e. column looks like:

food
["lettuce", "carrots"]
["carrots", "zucchini"]

just use the column name directly inside the brackets:

select * from rabbits where (food)::jsonb ? 'carrots';

Upvotes: 16

Sam Hughes
Sam Hughes

Reputation: 785

Not simpler but smarter:

select json_path_query(info, '$ ? (@.food[*] == "carrots")') from rabbits

Upvotes: 2

Snowball
Snowball

Reputation: 11696

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

Upvotes: 368

makasprzak
makasprzak

Reputation: 5220

A small variation but nothing new infact. It's really missing a feature...

select info->>'name' from rabbits 
where '"carrots"' = ANY (ARRAY(
    select * from json_array_elements(info->'food'))::text[]);

Upvotes: 20

gori
gori

Reputation: 1258

You could use @> operator to do this something like

SELECT info->>'name'
FROM rabbits
WHERE info->'food' @> '"carrots"';

Upvotes: 69

chrmod
chrmod

Reputation: 1445

Not smarter but simpler:

select info->>'name' from rabbits WHERE info->>'food' LIKE '%"carrots"%';

Upvotes: 31

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