Question

$stmt = $dbo->getConnection()->prepare("SELECT count(character_name)
            FROM playercharacter
            JOIN character
            ON (playercharacter.character_id = character.character_id)
            WHERE user_id = :user_id");

给我

致命错误：带有消息的未捕获异常'PDOException' 'SQLSTATE [42000]：语法错误或访问冲突：1064您有 SQL语法错误;查看与您的手册相对应的手册 MySQL服务器版本，用于在'字符ON附近使用正确的语法（playercharacter.character_id = character.character_id）'在第3行' /home/content/50/10835750/html/dogdays/game/private.php:43堆栈追踪：＃0 /home/content/50/10835750/html/dogdays/game/private.php(43）： PDOStatement-＆gt;执行（数组）＃1 {main}抛出第43行/home/content/50/10835750/html/dogdays/game/private.php

Answer 1

character是保留关键字，恰好是您的表格的名称。为了避免语法错误，应使用反引号转义表名。前，

SELECT  count(character_name) 
FROM    playercharacter JOIN `character` 
            ON (playercharacter.character_id = `character`.character_id) 
WHERE   user_id = :user_id

MySQL Reserved Keywords List

如果您有权更改表格，请更改保留关键字列表上不存在的表名称，以防止同一错误再次返回。

JOIN语句中的SQL语法错误

1 个答案: