Question

我为我的用户注册页面写了一个MYSQL程序，我已经编写了PHP部分来从那里发送数据到MySQL，它工作正常（尝试使用虚拟数据和检索）。但我的处理程序或交易存在某种问题我想

这是我的程序：

BEGIN
DECLARE unamec INT;
DECLARE emailc INT;
DECLARE m INT;
DECLARE msg VARCHAR(100);
DECLARE EXIT HANDLER FOR SQLEXCEPTION SET m=1;
    START TRANSACTION;
    SET autocommit=0;
    SELECT COUNT(*) INTO unamec FROM login WHERE uname=`@user`;
    IF unamec=0 THEN
        SELECT COUNT(*) INTO emailc FROM login WHERE email=@email;
        IF emailc=0 THEN
            INSERT INTO login (uname, hash, email, role) values(@user,MD5(@password),@email,'1');
            SET msg='Successfully Registered';
        ELSE
            SET msg='Email Already Exists';
        END IF;
    ELSE
        SET msg='Username Already Exists';
    END IF;
COMMIT;
END;
END;
IF m=1 THEN
ROLLBACK;
SET msg='ERROR';
END IF;
SELECT msg as message;
END

我总是收到错误

MySQL said: #1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near
'END;
IF m=1 THEN
ROLLBACK;
SET msg='ERROR';
END IF;
SELECT msg as message;' at line 23

有什么东西显而易见我错过了吗？我已经尝试了一段时间来解决这个问题，几乎所有的结果都显示出同样的错误。

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Answer 1

好的，我编辑了自己的代码并获得了所需的结果。

我所做的改变是：

BEGIN
DECLARE unamec INT;
DECLARE emailc INT;
DECLARE m INT;
DECLARE msg VARCHAR(100);
DECLARE EXIT HANDLER FOR SQLEXCEPTION
START TRANSACTION;
SET autocommit=0;
SELECT COUNT(*) INTO unamec FROM login WHERE uname=`@user`;
IF unamec=0 THEN
    SELECT COUNT(*) INTO emailc FROM login WHERE email=@email;
    IF emailc=0 THEN
        INSERT INTO login (uname, hash, email, role) values(@user,MD5(@password),@email,'1');
        SET msg='Successfully Registered';
    ELSE
        SET msg='Email Already Exists';
    END IF;
ELSE
    SET msg='Username Already Exists';
END IF;
COMMIT;
BEGIN
ROLLBACK;
SET msg='ERROR';
END;
SELECT msg as message;
END

如果有人得到任何其他答案，仍然会感激。

无论如何，上面的代码现在工作正常。

MySQL说1064你的SQL语法有错误

1 个答案: