Выберите элемент.* как json mysql
-- You can't use * in json selects in all version of mySQL, but you can
-- outline all fields you want in a select statement as follows:
SELECT contact.*, JSON_OBJECT(
'name_field', name_field,
'address_field', address_field,
'contact_age', contact_age
) AS location
FROM contact
JOIN location on contact.loc_id = location.id;
-- You may be able to use "JSON_ARRAYAGG()" for this in mySQL 8+
PDXfoster