问题
I'm still pretty new to handling JSON fields in MySQL. All the solutions I've come across deal with objects that have key/values; unable to find one that handles JSON arrays.
Anyways, what I want to do is to be able to select all rows where the interestIds contain 2 in them. How do I do that? Thanks.
Users table
+----+-------------+
| id | interestIds |
+----+-------------+
| 1 | [1, 2] |
| 2 | [3, 2] |
| 3 | [2, 4] |
+----+-------------+
Sample test query:
SET @userId = 2;
SELECT * FROM Users
WHERE @userId IN JSON_CONTAINS(@user, interestIds, '$[1]');
I am confused as how to use the JSON_* functions; not sure what to put for the 3rd parameter...
回答1:
You can use the following solution, using JSON_CONTAINS:
SELECT *
FROM Users
WHERE JSON_CONTAINS(interestIds, '2') = 1;
The third (optional) paramater path gives you the posibility to use this function only on a specific part of your JSON value. So the following example checks if 2 is the second value of the array:
SELECT *
FROM test
WHERE JSON_CONTAINS(interestIds, '2', '$[1]') = 1;
demo on dbfiddle.uk
回答2:
Use JSON_SEARCH which returns path to element you are searching, or null if not found:
SELECT *
FROM users
WHERE JSON_SEARCH(interestids, 'one', '2') IS NOT NULL
Live Demo
If you're storing many-to-many relationship using simple JSON array, there are better ways to do it. Consider creating user_interest table and doing it the right and simpler way. That is if your JSON actually looks like you have shown us and does not contain dynamic key-value pairs.
回答3:
SQL> select id
from users
where JSON_CONTAINS(interestIds, "2","$");
+----+
| id |
+----+
| 1 |
| 2 |
| 3 |
+----+
3 rows in set (0.0015 sec)
回答4:
Wrap your select to JSON_ARRAYAGG
Like:
SELECT JSON_ARRAYAGG(JSON_OBJECT(....)) FROM table....
来源：https://stackoverflow.com/questions/51873294/mysql-select-records-based-on-values-in-json-array