Using MySQL JSON field to join on a table

Question

I have a json field that stores a list of ids (not best practice here I know), I want to know if it's possible to use do operations on this JSON field and use them in the sql.

Below is a fictitious example of what I'm trying to achieve, is something like this doable?

CREATE TABLE user (
    user_id INT,
    user_name VARCHAR(50),
    user_groups JSON
);

CREATE TABLE user_group (
    user_group_id INT,
    group_name VARCHAR(50)
);

INSERT INTO user_group (user_group_id, group_name) VALUES (1, 'Group A');
INSERT INTO user_group (user_group_id, group_name) VALUES (2, 'Group B');
INSERT INTO user_group (user_group_id, group_name) VALUES (3, 'Group C');

INSERT INTO user (user_id, user_name, user_groups) VALUES (101, 'John', '[1,3]');

With the above data I would like to fashion a query that gives me the results like this:

user_id | user_name | user_group_id | group_name|
-------------------------------------------------
101     | John      | 1             | Group A
101     | John      | 3             | Group C

Some psuedo style SQL I'm thinking is below, though I still have no clue if this is possible, or what JSON functions mysql offers I would use to achieve this

 SELECT 
       u.user_id, 
       u.user_name, 
       g.user_group_id
       g.group_name
   FROM users u
   LEFT JOIN user_group g on g.user_group_id in some_json_function?(u.user_groups)

Let me know if the question isn't clear.

Answer 1

With the help of Feras's comment and some fiddling:

  SELECT 
       u.user_id, 
       u.user_name, 
       g.user_group_id,
       g.group_name
   FROM user u
   LEFT JOIN user_group g on JSON_CONTAINS(u.user_groups, CAST(g.user_group_id as JSON), '$')

This appears to work, let me know if there's a better way.

Answer 2

Funny, I got to the opposite solution compared to Kyle's.

I wrote my query like this:

SELECT 
       u.user_id, 
       u.user_name, 
       g.user_group_id,
       g.group_name
   FROM user u
   LEFT JOIN user_group g on JSON_UNQUOTE(JSON_EXTRACT(u.user_groups, '$')) = g.user_group_id;

It also works, and this solution doesn't need any transforming on the right side of the expression, this could provide a benefit in query optimizing in certain cases.