尝试使用联接时我的 SQL 语法出错

Question

我目前正在尝试执行此 sql 查询，但我在 SQL:

方面不是最出色的

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 = ?
`;

但是我得到这个错误：

[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 }

任何信息都将非常感谢。

Answer 1

Char是保留关键字，可以在https://dev.mysql.com/doc/refman/5.7/en/keywords.html.

中查找

所以将代码改成这样，就可以了。一般远离与数据类型冲突的变量名。

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

Answer 2

语法为 LEFT JOIN chars as 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 = ?

为什么你仍然想要姓名、经验和锁定的字段？

尝试使用联接时我的 SQL 语法出错

Having an error in my SQL syntax when trying to use joins

mysql

mariadb