Question

我有一个简单的简单问题！ :-D 我是Php的新手，我写了一个简单的code，因此我可以通过SSH连接到设备。但是当我在一个类和方法中尝试这个时，不起作用。这是我的主要Php的代码，也包括我的课程：

<?php
class Connection {
    public $ip;
    public $usernam;
    public $password;
    public  $ssh;

    public  function sshConnection() {

        include ('./view/login.html');
        include('Net/SSH2.php'); // I use phpseclib to connect via SSH
        if(isset($_POST['login'])) { // Login button in html file


            $this->ip = $_POST['ip']; // input type to get ip in html file
            $this->usernam = $_POST['username']; // input type to get username in html file
            $this->password = $_POST['password']; // input type to get password in html file
            $this->ssh = new Net_SSH2($ip);
            if (!$ssh->login($username, $password)) {

                print('Login faild');
            } else {
                echo $ssh->exec('pwd');

            }
        }
    }
}


$connection=new Connection();

$connection->sshConnection();

我所包含的html文件登录。html是包含内部CSS的登录页面的视图。如您所见，我已通过此代码创建了一个类：

$connection=new Connection();       
$connection->sshConnection();

当我使用此代码而不创建类时，它工作正常！当我定义类时它不起作用，按下Login Button后没有任何反应。

我做错了什么？任何帮助将不胜感激。： - ）

Answer 1

所以，正如我在评论中所说，你的代码应该是这样的（我添加了一点奖励 - isset（）可以在单个调用中检查很多变量）：

<?php
class Connection {
    public $ip;
    public $username;
    public $password;
    public $ssh;

    public  function sshConnection() {
        include ('./view/login.html');
        include('Net/SSH2.php'); // I use phpseclib to connect via SSH

        if(isset($_POST['login'], $_POST['ip'], $_POST['username'], $_POST['password'])) { // Login button in html file
            $this->ip = $_POST['ip']; // input type to get ip in html file
            $this->username = $_POST['username']; // input type to get username in html file
            $this->password = $_POST['password']; // input type to get password in html file
            $this->ssh = new Net_SSH2($this->ip);
            if (!$this->ssh->login($this->username, $this->password)) {
                print('Login faild');
            } else {
                echo $this->ssh->exec('pwd');

            }
        }
    }
}


$connection=new Connection();
$connection->sshConnection();

干杯！

Answer 2

感谢我的朋友@arbogastes，我的问题非常简单！我忘了为我定义的所有变量添加$ this。还要感谢@Echoes和@zenwraight。所以我的代码就这样了：

<?php
             class Connection {
             public $ip;
             public $usernam;
             public $password;
             public  $ssh;

            public  function sshConnection() {

                include ('./view/login.html');
                include('Net/SSH2.php'); // I use phpseclib to connect via SSH
                if(isset($_POST['login'])) { // Login button in html file



             $this->ip = $_POST['ip'];
             $this->usernam = $_POST['username'];
             $this->password = $_POST['password'];
             $this->ssh = new Net_SSH2($this->ip);
            if (!$this->ssh->login($this->username, $this->password)) {
            exit('Login Failed');
            print('Login faild');
        } else {
        echo $ssh->exec('pwd');

        }
         }
           }
         }

         $connection=new Connection();

         $connection->sshConnection();
?>

Answer 3

试试这个：

<?php
class Connection {
    public $ip;
    public $usernam;
    public $password;
    public  $ssh;

    public function sshConnection($ip, $username, $password) {
        include ('./view/login.html');
        include('Net/SSH2.php'); 

        $this->ip = $ip; // input type to get ip in html file
        $this->usernam = $username; // input type to get username in html file
        $this->password = $password; // input type to get password in html file
        $this->ssh = new Net_SSH2($this->ip); 
        if (!$this->ssh->login($this->usernam, $this->password)) {
            print('Login faild');
        } 
        else {
            echo $this->ssh->exec('pwd');
        }
    }
}

if(isset($_POST['login'])) {
    $connection=new Connection();
    $connection->sshConnection($_POST['ip'], $_POST['username'], $_POST['password']);
}

?>

无法调用PHP类的方法

3 个答案: