Jump to content

mysql error


justinh

Recommended Posts

My login form is processed using processlogin.php

 

you can see what goes wrong here:

 

www.wmptest.com/stbs/index.html

 

 

Error is:

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /homepages/29/d119570661/htdocs/wmptest.com/stbs/processlogin.php on line 82

 

 

 

<?php
session_start();


include("connect.php");

$username = stripslashes($_POST['username']);
$password = stripslashes($_POST['password']);


if (!$username || !$password){

              ?>                         <html>
  <head>
   <title> South Texas Business Systems - UploadManager </title>
   <link rel="stylesheet" type="text/css" href="style.css">
   <script type="text/javascript">
function showPopup(url) {
newwindow=window.open(url,'name','height=190,width=520,top=320,left=100,resizable,scrollable');
if (window.focus) {newwindow.focus()}
}
</script>
  </head>
  <body>
  <div id="container">
  <img src="uploadmanager.jpg"></img><img src="uploadmanager2.jpg"></img><div id="content">
  <form method="POST" action="processlogin.php">
  <fieldset>
  <font color="red"><small>Must provide a username and password</small><Br></font>
  <legend> Login<a href='help.html' onClick='showPopup(this.href);return(false);'>[?]</a>  </legend>
  <label for="username">User Name:</label><input type="text" name="username">
  <label for="password">Password:</label><input type="password" name="password">
  <input type="submit" class="submitbutton" value="Login"><br><div id="copyright">Copyright 2008-2009&#169; 2008-2009, All rights reserved. This program is protected by local and international copyright laws as described in the license agreement.</div>
  </fieldset>
  </form>

  </div>
  </div>
  </body>
  </html> <?


}else{

$finduser = mysql_query("SELECT * FROM tbl_users WHERE username = $username AND password = $password");


if(!finduser){
?>  <html>
  <head>
   <title> South Texas Business Systems - UploadManager </title>
   <link rel="stylesheet" type="text/css" href="style.css">
   <script type="text/javascript">
function showPopup(url) {
newwindow=window.open(url,'name','height=190,width=520,top=320,left=100,resizable,scrollable');
if (window.focus) {newwindow.focus()}
}
</script>
  </head>
  <body>
  <div id="container">
  <img src="uploadmanager.jpg"></img><img src="uploadmanager2.jpg"></img><div id="content">
  <form method="POST" action="processlogin.php">
  <fieldset>
  <font color="red"><small>User name or passsword is incorrect.</small><Br></font>
  <legend> Login<a href='help.html' onClick='showPopup(this.href);return(false);'>[?]</a>  </legend>
  <label for="username">User Name:</label><input type="text" name="username">
  <label for="password">Password:</label><input type="password" name="password">
  <input type="submit" class="submitbutton" value="Login"><br><div id="copyright">Copyright 2008-2009&#169; 2008-2009, All rights reserved. This program is protected by local and international copyright laws as described in the license agreement.</div>
  </fieldset>
  </form>

  </div>
  </div>
  </body>
  </html>
  
  <?
    
    }else{

    while($getuserinfo = mysql_fetch_array($finduser)){
    $username = $getuserinfo['username'];
    $rank = $getuserinfo['siterank'];
    $firstname = $getuserinfo['firstname'];
    }
    
    $_SESSION['loggedin'] = 1;
    $_SESSION['siterank'] = $rank;
    $_SESSION['firstname'] = $firstname;
    
    
   ?>
    <html>
<head>
<title>
South Texas Business Systems - UploadManager</title>
<link rel="stylesheet" type="text/css" href="style.css">
</head>
<body>
<div id="container"><img src="uploadmanager.jpg"></img><img src="uploadmanager2.jpg"></img><div id="content2">
<br><br><br><b>Log in successful!</b> <br><img src="ajax-loader.gif" class="loading"></img>Now redirecting you to your account...<meta http-equiv="Refresh" content="5; url=about:blank">
</div>
</div>
</body>
</html>

    
<?   } }   ?>

Link to comment
https://forums.phpfreaks.com/topic/136746-mysql-error/
Share on other sites

change

 

$finduser = mysql_query("SELECT * FROM tbl_users WHERE username = $username AND password = $password");

 

to

 

$finduser = mysql_query("SELECT * FROM tbl_users WHERE username = '$username' AND password = '$password'") or die("Error:".mysql_error()." No:".mysql_errno());

Link to comment
https://forums.phpfreaks.com/topic/136746-mysql-error/#findComment-714175
Share on other sites

Archived

This topic is now archived and is closed to further replies.

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.