我确定这个问题之前已经被问过,但我已经彻底寻找答案但无济于事。(我见过的唯一答案涉及 ajax)但我只使用 javascript、PHP 和 HTML。
我有一个 login.php 页面,并且我已经创建了一个 HTML 页面,该页面将在用户成功登录后立即成为登录页面。我该怎么做?
下面是我的登录页面的代码,登录后的登陆页面叫做transfer.html:
登录.PHP
<div id="content">
<h3>Login to Internet Banking</h3>
<form id="login" action="" method="post">
<p>
<label for="userid">UserID:</label>
<input type="text" name="UserID" id="UserID"/>
</p>
<p>
<label for="PIN">PIN:</label>
<input type="password" name="PIN" id="PIN" />
</p>
<p>
<input type="submit" name="btnSend" value="Login" class="submit_button" />
</p>
</form>
<td> </td>
<p>
Not yet registered?
<a href="registration.php">Click here to register</a>
</p>
<div id="wrap">
<!-- start PHP code -->
<?php
mysql_connect("localhost", "root", "") or die(mysql_error()); // Connect to database server(localhost) with UserID and PIN.
mysql_select_db("registrations") or die(mysql_error()); // Select registration database.
if(isset($_POST['name']) && !empty($_POST['name']) AND isset($_POST['PIN']) && !empty($_POST['PIN'])){
$UserID = mysql_escape_string($_POST['name']);
$PIN = mysql_escape_string(md5($_POST['PIN']));
$search = mysql_query("SELECT UserID, PIN, active FROM users WHERE UserID='".$UserID."' AND PIN='".$PIN."' AND active='1'") or die(mysql_error());
$match = mysql_num_rows($search);
if($match > 0){
$msg = 'Login Complete! Thanks';
}else{
$msg = 'Login Failed!<br /> Please make sure that you enter the correct details and that you have activated your account.';
}
}
?>
<!-- stop PHP Code -->
<?php
if(isset($msg)){ // Check if $msg is not empty
echo '<div class="statusmsg">'.$msg.'</div>'; // Display our message and add a div around it with the class statusmsg
} ?>
</div>
</div>