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您好，我无法理解为什么我的身份验证不断失败。我尝试了许多调试方法，我想我很难过。请记住，我仍然是初学者，目前安全性不是问题。我只是想让它工作。

所以对于我的注册脚本，我让用户输入用户名和密码

            $username  = $_POST['username'];
            $password  = $_POST['password'];

// Post 正在保存的数据

            $createTable = $db->prepare("CREATE TABLE `$username` (Title varchar(40) ,
                                                    About   varchar(1000) , 
                                                    Chapter int(4),
                                                    Img varchar(100) , 
                                                    Url varchar(50)) ");

            $createTable->execute();

/// 为用户创建一个表

            $Register = $db->prepare("INSERT INTO library (id, 

username, password) VALUES ($id ,
'$username','$password')");
            $Register ->execute();

// 将数据插入到库表中。图书馆有用户和密码

                echo 'user sucessfully registered';

                $_SESSION['user'] = $username;
                $_SESSION['pass'] = $password;

我遇到的问题是对此进行身份验证以进行登录。使用两个 SESSION 变量作为参数调用登录函数。

function login($SentUser , $SentPass)

      {

echo 'trying to log in '.'<br>';

    require_once 'Connection.php';
    $user = $SentUser;
    $pass = $SentPass;

    /// check data base for match user and pass
        $sth = $db->prepare('SELECT username , password from `library` where username = ? And password = ? ');

        $sth->bindParam(1, $user);
        $sth->bindParam(2, $pass);
        $sth->execute();


        while ($row = $sth->fetch(PDO::FETCH_ASSOC))
            {
        echo 'user : ' . $row['username'] , '<br>';
        echo 'pass: ' . $row['password'] , '<br>';
        echo '<br>';
            }
        .
        // fetchcolumn returns true false if select took a row or not.
if($sth->fetchColumn() != false) { /// This is where the code fails as it almost always goes to the else case.

            $_SESSION['user'] = $user;
            $_SESSION['pass'] = $pass;
            $_SESSION['passed'] = TRUE;
            echo 'Welcome '.$user.'<br>';}
        else
            {

                echo 'failed authenticate'.'<br>';

            }

为了更加清楚，这里是被调用的connection.php。它基本上只是将我连接到数据库，为我提供 $db 对象。

<?php


$config['db'] = array(
    'host' => 'localhost',
    'username' => 'root',
    'password' => '',
    'dbname' => 'mywebsite'
    );




$db = new PDO('mysql:host=' . $config['db']['host']. ';dbname=' . $config['db']['dbname'] , $config['db']['username'] , $config['db']['password']);

?>
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1 回答 1

1

你提到library的表有用户/密码。因此，每个用户必须只有一行。更改代码如下：

   if($sth->rowCount() == 1) {
      //row exists, so user exists
      $row = $sth->fetch(PDO::FETCH_ASSOC);
      echo 'user : ' . $row['username'] , '<br>';
      echo 'pass: ' . $row['password'] , '<br>';
      echo '<br>';

      //set session
      $_SESSION['user'] = $user;
      $_SESSION['pass'] = $pass;
      $_SESSION['passed'] = TRUE;
      echo 'Welcome '.$user.'<br>';
   } else {
      //auth failed
      echo 'failed authenticate'.'<br>';
   }
于 2013-10-05T08:44:21.420 回答