Question

所以我一直在学习如何制作一个C＃应用程序，最近用LAMP服务器发送和接收参数但是我似乎已经碰壁了，我需要一些帮助来理解我在这里做错了什么。

这是一个简单的测试应用程序，它将两个参数（登录名和密码）发送到php文件，并根据参数是否正确接收响应。

private void btEntrar_Click(object sender, EventArgs e)
    {
        String login, senha, postData, requestMethod = "POST", contentType = "application/x-www-form-urlencoded", responseFromServer = null;
        String urlLogin = "http://192.168.1.107/knowledgems/accounts/login.php";

        login = tbLogin.Text;
        senha = tbSenha.Text;
        postData = "param1=" + login + "&param2=" + senha;
        byte[] byteArray = Encoding.UTF8.GetBytes(postData);
        WebRequest request = WebRequest.Create(urlLogin);
        WebResponse response;
        Stream dataStream;
        StreamReader reader;

        //Send request
        request.Method = requestMethod;
        request.ContentType = contentType;
        request.ContentLength = byteArray.Length;
        dataStream = request.GetRequestStream();
        dataStream.Write(byteArray, 0, byteArray.Length);
        dataStream.Close();

        //Get response
        response = request.GetResponse();
        dataStream = response.GetResponseStream();
        reader = new StreamReader(dataStream);
        responseFromServer = reader.ReadToEnd();
        reader.Close();
        dataStream.Close();
        response.Close();

        if (responseFromServer.Equals("SUCCESS"))
            MessageBox.Show("Login efetuado.");
        else
            MessageBox.Show("Login não efetuado." + responseFromServer);
    } //btEntrar

虽然一切看起来都是正确的，事实上我可以从我的php文件中得到回复，但我最终无法通过验证功能

(if (responseFromServer.Equals("SUCCESS")))

即使显示的响应正好是字符串“SUCCESS”，我似乎无法将responseFromServer字符串与“SUCCESS”进行比较，以检查是否一切都可以登录，因此它总是最终会读取else条件。

以下我也发布了我的php文件。

<?php
require_once 'connection.php';
header('Content-Type: application/form-data');

class User {
    private $db;
    private $connection;

    function __construct(){
        $this->db = new DB_Connection();
        $this->connection = $this->db->get_connection();            
    } //__construct

    public function does_user_exist($username,$password){
        $query = "SELECT * FROM users WHERE user_username = '$username' AND user_password = '$password'";
        $result = mysqli_query($this->connection,$query);
        if(mysqli_num_rows($result) > 0){
            echo "SUCCESS";     
        } //if
        else {
            echo "ERRO&101";            
        } //else            
        mysqli_close($this->connection);
    } //does_user_exist     

} //class

$user = new User();
if (isset($_POST['param1'],$_POST['param2'])){
    $username = $_POST['param1'];
    $password = $_POST['param2'];
    if(!empty($username) && !empty($password)){
        //$encrypted_password = md5($password);
        $user -> does_user_exist($username,$password);
    }
    else {
        echo "ERRO&100";
    } //else
} //if
?>

感谢您提前的时间，请帮忙！

Edit1：我的猜测是这种格式对于contentType来说不太合适，但我无法弄清楚这是什么样的正确方法。

Answer 1

“成功”可能会在最后换行吗？看看是否有效：

if (responseFromServer.Trim().Equals("SUCCESS"))
{

如何使此代码正常工作？

1 个答案: