assign json array (id's) with mysql tables PHP [duplicate] - php

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.

Answer 3

For arrays like ["1", "2", "3"] that values are in string type, JSON_SEARCH function is the way for your question:
SELECT
u.user_id,
u.user_name,
g.user_group_id
g.group_name
FROM users u
LEFT JOIN user_group g ON (JSON_SEARCH(u.user_groups, 'one', g.user_group_id))
JSON_CONTAINS function does not return true for integers as candidate parameter:
SELECT JSON_CONTAINS(CAST('["1", "2", "3"]' AS JSON), CAST(1 AS JSON), '$')
returns 0 (false). You need to change it to this:
SELECT JSON_CONTAINS(CAST('["1", "2", "3"]' AS JSON), CAST(CONCAT('"', 1, '"') AS JSON), '$')
But JSON_SEARCH can find the result:
SELECT JSON_SEARCH(CAST('["1", "2", "3"]' AS JSON), 'one', 1)
returns "$[0]" that means "true".

Answer 4

I have just tried the following and it worked in mysql 8.0.26:
-- index
ALTER TABLE user ADD KEY ( (CAST(user_groups -> '$' AS UNSIGNED ARRAY)) );
SELECT *
FROM user as a
left join user_group as t on t.user_group_id MEMBER OF (a.user_groups)
where t.group_name in ('Group A', 'Group C');
The query is based on the following select syntax:
SELECT * FROM t_json_arrays WHERE 3 MEMBER OF (c_array);
See examples on: https://saveriomiroddi.github.io/Storage-and-indexed-access-of-denormalized-columns-arrays-on-mysql-8.0-via-multi-valued-indexes/