Question

我不知道如何处理这个错误我的代码

Notice: Use of undefined constant result2 - assumed 'result2' in C:\xampp\htdocs\how are things\admin panel\register3.php on line 62

第62行的代码是

 $result2 = mysql_query("INSERT INTO users (id ,username, user_level, type, first_name, last_name, email, password, phone_number) VALUES('','$username', '2', 'a','$first_name', '$last_name', '$email','$password','$phone_number') ") or die(mysql_error());

这是找到它的代码

if (empty($error)){
     $result = mysql_query("SELECT * FROM users WHERE email='$email' OR username='$username' ") or die(mysql_error());
     $result2 = mysql_query("INSERT INTO users (id ,username, user_level, type, first_name, last_name, email, password, phone_number) VALUES('','$username', '2', 'a','$first_name', '$last_name', '$email','$password','$phone_number') ") or die(mysql_error());
     if(!result2){
        die('Could not insert into the database: '.mysql_error());
     }

    } else{
    $error_message = '<span class="error">';
    foreach($error as $key => $Values){
    $error_message.= "$Values";

    }
    $error_message.="</span><br/><br/>";

    }

Answer 1

if(!result2){

你在那里错过了$。请使用IDE进行编码，这样就可以避免这种错误

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