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Editorial Team
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Asked: 2026-06-06T11:56:21+00:00

I have a php file which checks for login and password from users database,

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I have a php file which checks for login and password from users database, it works fine.

But I am unable to validate the exact error to display if the user does not exist or password incorrect to the user and go back to previous page after error, help me how to display these errors.

<?php // access.php
include_once 'common.php';
include_once 'db.php';

session_start();

$uid = isset($_POST['uid']) ? $_POST['uid'] : $_SESSION['uid'];
$pwd = isset($_POST['pwd']) ? $_POST['pwd'] : $_SESSION['pwd'];

if(!isset($uid)) {
  ?>

  <!DOCTYPE html PUBLIC "-//W3C/DTD XHTML 1.0 Transitional//EN" 
    "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
  <html xmlns="http://www.w3.org/1999/xhtml">
  <head>
<title>Login</title>
    <meta http-equiv="Content-Type"
      content="text/html; charset=iso-8859-1" />
<head>
<style type="text/css">
<!--
.style1 {
    font-size: 16px;
    font-family: Verdana, Arial, Helvetica, sans-serif;
}
.style3 {
    font-size: 12px;
    font-family: Verdana, Arial, Helvetica, sans-serif;
}
-->
</style>

  </head>
<body>
  <h1 class="style1"> <br><br>  Login Required </h1>
  <span class="style3"><br>
  You <strong>must login to access this area </strong>of the site. <br>
  <br>
  If you are not a registered user, please contact your Admin
     to sign up for instant access!</span>
  <p><form method="post" action="<?=$_SERVER['PHP_SELF']?>">
    <span class="style3">User ID:&nbsp;&nbsp;&nbsp;&nbsp;    
    <input type="text" name="uid" size="12" />
    <br>
    <br />
    Password:</span>    
    <input type="password" name="pwd" SIZE="12" />
    <br>
    <br />
    <input type="submit" value="Login" />
  </form></p>
</body>
 </html>

  <?php
  exit;
}

$_SESSION['uid'] = $uid;
$_SESSION['pwd'] = $pwd;

dbConnect("svga");
$sql = "SELECT * FROM user WHERE
        userid = '$uid' AND password = '$pwd'";
$result = mysql_query($sql);
if (!$result) {
  error('A database error occurred while checking your '.
        'login details.\\nIf this error persists, please '.
        'contact you@example.com.');
   }

if (mysql_num_rows($result) == 0) {
  unset($_SESSION['uid']);
  unset($_SESSION['pwd']);
  ?>

  <!DOCTYPE html PUBLIC "-//W3C/DTD XHTML 1.0 Transitional//EN"
    "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
  <html xmlns="http://www.w3.org/1999/xhtml">
      <head>
    <title> Access Denied </title>
    <meta http-equiv="Content-Type"
      content="text/html; charset=iso-8859-1" />
        <style type="text/css">
<!--
.style1 {
    font-size: 16px;
    font-family: Verdana, Arial, Helvetica, sans-serif;
}
.style3 {
    font-size: 12px;
    font-family: Verdana, Arial, Helvetica, sans-serif;
}
-->
</style>  

  </head>
  <body>
  <br/>
  <br/>

  <h1 class="style1"> Access Denied </h1>
  <p class="style3">Your user ID or password is incorrect, or you are not a
     registered user on this site. To try logging in again, click
     <a href="<?=$_SERVER['PHP_SELF']?>">here</a>. To access, please contact our Admin     !</a>.</p>
  </body>
  </html>

<?php
  exit;
}
$username = mysql_result($result,0,'fullname');
$_SESSION['user'] = mysql_result($result,0,'userid');
?>
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1 Answer

  1. Editorial Team

    simply put

    if(mysql_num_rows($result)==0) {
header('location:your_page.php');
}

    this will redirect you to the page and assuming that you have defined error() method, just 

    echo error();
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