Question

我有一个网站应该有一个论坛，你登录。不幸的是它没有出现。有2个文件：index.php，另一个是connect.php。我试图从index.php中删除php代码，该代码应该查看用户名和密码是否混淆了数据库中的内容并显示基于此的消息。然后论坛出现了。然后我把它加回来它就消失了。我什么都不知道，请帮助我。

的index.php：

<?php
session_start();
    if(isset($msg) & !empty($msg)){
        echo $msg;
    }
    ?>

<div class="register-form">
<h1>Login</h1>
<form action="" method="POST">
    <p><label>User Name : </label>
    <input id="username" type="text" name="username" placeholder="username" /></p>

     <p><label>Password&nbsp;&nbsp; : </label>
     <input id="password" type="password" name="password" placeholder="password" /></p>

    <a class="btn" href="register.php">Signup</a>
    <input class="btn register" type="submit" name="submit" value="Login" />
    </form>
</div>

<?php
  //Start the Session

 require('connect.php');
//3. If the form is submitted or not.
//3.1 If the form is submitted
if (isset($_POST['username']) and isset($_POST['password'])){
//3.1.1 Assigning posted values to variables.
$username = $_POST['username'];
$password = $_POST['password'];
//3.1.2 Checking the values are existing in the database or not
$query = "SELECT * FROM `user` WHERE username='$username' and password='$password'";

$result = mysql_query($query) or die(mysql_error());
$count = mysql_num_rows($result);
//3.1.2 If the posted values are equal to the database values, then session will be created for the user.
if ($count == 1){
$_SESSION['username'] = $username;
}else{
//3.1.3 If the login credentials doesn't match, he will be shown with an error message.
echo "Invalid Login Credentials.";
}
}
//3.1.4 if the user is logged in Greets the user with message
if (isset($_SESSION['username'])){
$username = $_SESSION['username'];
echo "Hai " . $username . "
";
echo "This is the Members Area
";
echo "<a href='logout.php'>Logout</a>";

}else{
//3.2 When the user visits the page first time, simple login form will be displayed.
?>

connect.php：

<?php
$connection = mysql_connect('localhost', 'root', 'root');
if (!$connection){
    die("Database Connection Failed" . mysql_error());
}
$select_db = mysql_select_db('test');
if (!$select_db){
    die("Database Selection Failed" . mysql_error());
}

?>

Answer 1

你错过了前一行的最后一行？＆gt;
变化：

}else{
    //3.2 When the user visits the page first time, simple login form will be displayed.
?>

进入这个：

}else{
    //3.2 When the user visits the page first time, simple login form will be displayed.
}
?>

登录论坛没有显示

1 个答案: