Mysql的JSON路径提取

Question

表：相关信息

id  | info 
------ | ------ 
1  | {"v1": "test", "v2": "work"} 

表：my_data

id  | name  | info_id 
------ | ------ | ------ 
1  | john  | 2 
2  | peter | 1 
3  | luc  | 2 

我想有以下输出：

name  | art 
------ | ------ 
john  | work 
peter | test 
luc  | work 

我的问题是我不知道如何使用主查询的值作为json路径。

现在，我有这样的：

SELECT a.name, (SELECT info->>"$.v1" AS art FROM infos AS b) FROM my_data AS a 

可以肯定这显示了所有 “测试”

的查询必须是这样的：

SELECT a.name, (SELECT info->>CONCAT("$.v", a.info_id) AS art FROM infos AS b) FROM my_data AS a 

这可能吗？

Answer 1

你可以试试以下（根据需要调整）：

mysql> SELECT VERSION(); 
+-----------+ 
| VERSION() | 
+-----------+ 
| 5.7.19 | 
+-----------+ 
1 row in set (0.00 sec) 

mysql> DROP TABLE IF EXISTS `my_data`, `infos`; 
Query OK, 0 rows affected, 2 warnings (0.00 sec) 

mysql> CREATE TABLE IF NOT EXISTS `infos` (
    -> `id` BIGINT UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY, 
    -> `info` JSON NOT NULL 
    ->); 
Query OK, 0 rows affected (0.00 sec) 

mysql> CREATE TABLE IF NOT EXISTS `my_data` (
    -> `id` BIGINT UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY, 
    -> `name` VARCHAR(255) NOT NULL, 
    -> `info_id` BIGINT UNSIGNED NOT NULL 
    ->); 
Query OK, 0 rows affected (0.00 sec) 

mysql> INSERT INTO `infos` 
    -> (`info`) 
    -> VALUES 
    -> ('{"v1": "test", "v2": "work"}'); 
Query OK, 1 row affected (0.00 sec) 

mysql> INSERT INTO `my_data` 
    -> (`name`, `info_id`) 
    -> VALUES 
    -> ('john', 2), 
    -> ('peter', 1), 
    -> ('luc', 2); 
Query OK, 3 rows affected (0.00 sec) 
Records: 3 Duplicates: 0 Warnings: 0 

mysql> SELECT 
    -> `md`.`name`, 
    -> JSON_UNQUOTE(
    ->  JSON_EXTRACT(
    ->  `i`.`info`, 
    ->  CONCAT('$.v', `md`.`info_id`) 
    -> ) 
    -> ) `art` 
    -> FROM 
    -> `my_data` `md` 
    -> INNER JOIN `infos` `i` ON `i`.`id` = 1; 
+-------+------+ 
| name | art | 
+-------+------+ 
| john | work | 
| peter | test | 
| luc | work | 
+-------+------+ 
3 rows in set (0.00 sec) 

见db-fiddle。