尝试使用时有我的SQL语法错误连接

Question

我目前正在试图做到这一点sql查询，但我不是最大的SQL：

var sqlQuery = ` 
    SELECT user.ID, user.email, user.first_name, user.last_name, 
    user.address1, user.address2, user.country, user.gender 
    FROM subscriber AS user 
    LEFT JOIN (
     SELECT sID, name, experience, locked 
     FROM chars 
    ) AS char 
     ON user.ID = char.sID 
    WHERE user.ID = ? 
`; 

但我得到这个错误：

[1] Errors { [Error: ER_PARSE_ERROR: You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'char 
[1]   ON user.ID = char.sID 
[1]   WHERE user.ID = 13' at line 4] 
[1] code: 'ER_PARSE_ERROR', 
[1] errno: 1064, 
[1] sqlState: '42000', 
[1] index: 0 } 

任何信息将是非常感谢。

Answer 1

Char是一个保留关键字，可以在​​查询。

因此，将代码更改为这个，就是诀窍。一般来说，远离与数据类型冲突的变量名称。

(SELECT sID, name, experience, locked 
FROM chars) AS c 

Answer 2

语法将LEFT JOIN字符为char

var sqlQuery = ` 
SELECT user.ID, user.email, user.first_name, user.last_name, 
user.address1, user.address2, user.country, user.gender 
FROM subscriber AS user 
LEFT JOIN chars AS char 
    ON user.ID = char.sID 
WHERE user.ID = ? 

为什么你想要的名称，经验反正锁定领域？