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方法

JSON_CONTAINS(target, candidate[, path])

value MEMBER OF(json_array)

查询示例

mysql> set @list = JSON_ARRAY(1, 2);
Query OK, 0 rows affected (0.01 sec)

mysql> select 1 MEMBER OF(@list);
+--------------------+
| 1 MEMBER OF(@list) |
+--------------------+
| 1 |
+--------------------+
1 row in set (0.00 sec)


mysql> SELECT JSON_CONTAINS(@list, '1');
+---------------------------+
| JSON_CONTAINS(@list, '1') |
+---------------------------+
| 1 |
+---------------------------+
1 row in set (0.01 sec)