Question

我正在网站上工作。但是，我有一个问题，当我填写我的登录表单时，它会卡在加载页面上。并且似乎没有在用户列表中搜索注册用户。

我是新手，请耐心等待。

以下是LoginForm.php

<form name="login" method="post" action="login.php" onsubmit="return 
   validate();" >
   <div>Username: </br><input type="text" name="user" /> </div>
   </br>
   <div>Password:  </br><input type="password" name="pass" /> </div>
   </br>
   <div><input type="submit" value="Login"  style="background-color: #1daa87; 
      border: none; color: black; padding: 8px 52px;"></input> <input 
      type="reset" value="Reset" style="background-color: #1daa87; border: none; 
      color: black; padding: 8px 52px;"></input></div>
</form>

以下是login.php

    <?php
include_once ("dbconnection.php");

session_start(); //always start a session in the beginning

if ($_SERVER['REQUEST_METHOD'] == 'GET')
{
    if (empty($_GET['user']) || empty($_GET['pass'])) //Validating inputs using
    PHPcode
    {
    echo "Incorrect username or password";

    // header("location: LoginForm.php");//You will be sent to Login.php for re-

    login
    }

    $userName = $_GET["user"]; // as the method type in the form is "post" we
    areusing $_POSTotherwiseitwouldbe $_GET[]$password = $_GET["pass"];
    $stmt = $db->prepare("SELECT USER, PASS FROM WEBSITEUSERS WHERE USER = ?");

    // Fetching all the records with input credentials

    $stmt->bind_param("s", $username); //You need to specify values to each '?'
    explicitly
    while usingpreparedstatements $stmt->execute();
    $stmt->bind_result($UsernameDB, $PasswordDB); // Binding i.e. mapping
    databaseresultstonew variables

    // Compare if the database has username and password entered by the user.

    Passwordhastobedecrpted
    while comparing .
    if ($stmt->fetch() && password_verify($password, $PasswordDB))
        {
        $_SESSION['user'] = $userName; //Storing the username value in session
        variablesothatitcanberetrievedonotherpagesheader("location: userprofile.php"); // user will be taken to profile page
        }
      else
        {
        echo "<br />Incorrect username or password";
?>
    <br /><a href="LoginForm.php">Login</a>
    <?php
        }
}

?>

下面是dbconnection.php

<?php
//Establishing connection with the database
    define('DB_HOST', 'localhost'); 
    define('DB_NAME', ''); 
    define('DB_USER', ''); 
    define('DB_PASSWORD', '');  
    define('DB_DATABASE', 'WebsiteUsers');
$db = mysqli_connect(DB_HOST,DB_NAME,DB_PASSWORD,DB_USER);
?>

以下是userprofile.php

<?php
session_start();
$username = $_SESSION['user']; //retrieve the session variable

?>
    <center><h1>User Profile </h1></center>
    <br/>
    <b>Welcome <?php
echo $user ?> </b>
    <div style="text-align: right"><a href="logout.php">Logout</a></div> <!-- 
    calling Logout.php to destroy the session -->
    <?php

if (!isset($_SESSION['user'])) //If user is not logged in then he cannot
accesstheprofilepage
    {

    // echo 'You are not logged in. <a href="login.php">Click here</a> to log

    in .;
    header("location:LoginForm.php");
    }

?>

您可以尝试[test.laveshpillay.com/LoginForm.php] [1]的登录表单

您可以输入任何值，它会将您带到加载屏幕而不进行验证，它将保留在 login.php 表单

如果您需要代码的任何其他部分，请告诉我。

Answer 1

更改login.php文件中的以下内容并尝试：

<?php
if ($_SERVER['REQUEST_METHOD'] == 'POST'){//Made change here   
  use `$_REQUEST` instead of `$_GET`.
  //Your logic comes here.
}
?>

登录页面无效，html

1 个答案: