PHP 预处理语句找不到错误

Question

您好，我需要一个功能来检查用户是否已经存在于我的数据库中。我出于特定原因需要这个。此时函数始终returns false。有人看到我在代码中的错误了吗？

function checkForUserExists($username){
    $mysqli = new mysqli("****", "****", "****", "****");

        if($stmt = $mysqli->prepare("SELECT * FROM users WHERE ign = ?")){
        $stmt->bind_param("s", $username);        
        }else{

        }

        $stmt->execute();

        $stmt->bind_result($ign);

    if($ign == $username){
        return false;
    }else{
        return true;
    }
    $stmt->close();

}

更新

    $mysqli = new mysqli("**", "*", "**", "**");
    $username = "Felipebuy";
   if( $username && $stmt = $mysqli->prepare("SELECT ign FROM `users` WHERE `ign` = ?")){
        $stmt->bind_param("s", $username);
        $stmt->execute();
        $stmt->bind_result($ign);   
        $stmt->close();

       if($ign == $username){

        echo "equals";
       }else{
           echo "does not equal eachoter";
       }
    }

它只回显了"does not equal eachoter"

Answer 1

您在声明中 select ALL ( * )。除非您的 table 只包含一列，否则这将输出多个值。如果您将该结果 ($ign) 与单个值 ($username) 进行比较，它很可能永远不会 return true。

Answer 2

稍作更正，抱歉延迟 - 不得不睡觉。

    function checkForUserExists( $username=false ){
        $mysqli = new mysqli("****", "****", "****", "****");

        if( $username && $stmt = $mysqli->prepare("SELECT `ign` FROM `users` WHERE `ign` = ?")){
            $stmt->bind_param( "s", $username );
            $res=$stmt->execute();

            if( $res ){
                $stmt->bind_result( $ign );
                $stmt->fetch();  
                $stmt->close();
                $mysqli->close();

                return $ign === $username;
            }
        }
        return false;
    }

    $result=call_user_func('checkForUserExists','Felipebuy');
    echo $result ? 'equals' : 'not equal to each other';

PHP 预处理语句找不到错误

PHP prepared statement cant't find mistake

php

mysqli

prepared-statement