PHP - How To Get Details Of An Unregistred User ???

Actually, what i want to do is to use the email to fetch the $email,$password and $randomnumber from database after

Hello everyone.   I have a fully working form that gets data from a user (with $_post array) , and stores it in a database (mysql).   After successfulIy filling the form, I refer him to the "dashboard" page. In this page, i am having trouble to get his details from the database. How should I recognize him as the user that just registered?   should I use a $_post? or maybe a session? could you please give me a clue how to solve this? 

after the user has logged in, I would like to display their details by barcode id


Login.php

<?php
$host=""; // Host name
$username=""; // Mysql username
$password=""; // Mysql password
$db_name=""; // Database name
$tbl_name=""; // Table name

// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");



session_start();
// username and password sent from form
$barcodeID=$_POST['barcode'];



// To protect MySQL injection (more detail about MySQL injection)
$barcodeID = stripslashes($barcodeID);
$barcodeID = mysql_real_escape_string($barcodeID);


$sql="SELECT * FROM $tbl_name WHERE BarcodeID='$barcodeID'";


$result=mysql_query($sql);
     
$count=mysql_num_rows($result);

if($count > 0){
      $data = mysql_fetch_array ($result);
      
       $_SESSION["user_id"] = $data["BarcodeID"];
        $_SESSION["user_firstname"] = $data["Firstname"];
        $_SESSION["user_surname"] = $data["Surname"];
        $_SESSION["user_jobrole"] = $data["JobRole"];
        $_SESSION["user_manager"] = $data["Manager"];
        $_SESSION["user_priority"] = $data["Priority"];
        $_SESSION["user_datejoined"] = $data["DateJoined"];
        $_SESSION["user_times_loggged_in"] = $data["TimesLoggedOn"];

        if ($_SESSION["user_priority"] == '1')
        {
            header("Location: AdminSection.php");           
        }
        else
        {
            header("Location:LoggedIn.php");
      }      

      if ($_SESSION["user_times_loggged_in"] == '0')
      {
         
         header("Location:UsingTheSystem.html");
      }
}
         
?>



LoggedIn.php



I keep getting the error undefined index "barcode"?

<?php

$barcodeID = $_POST["barcode"];


include 'dbcon.php';

    $sql = "SELECT Firstname, Surname, JobRole, Manager"
        . " FROM users"
        . " WHERE BarcodeID = .'$barcodeID'" ;
    $rows = mysql_query($sql);
   
    echo $rows;
   
    ?>


Any help will be greatly appreciated


Thanks

I have created a button which when pressed should present the user with their details (whoever is logged in), here is the form code:

<form id="form1" name="form1" method="post" action="getdetails.php">
 
  <input type="submit" name="Get Details" value="Get Details" />
   </label>
   
  </p>
</form>

Here is the getdetails.php file

<?php
 mysql_connect("localhost","root","");
 mysql_select_db("test");
 $username = $_POST['textfield'];
  echo '</br>';
 $query = mysql_query("SELECT * FROM membersdetails WHERE name=`$username` ");

while($result = mysql_fetch_array($query)) {
//display
echo $result['firstname'];
echo $result['surname'];
}
?>

Its not workin at all I have attacthed the error i am getting

Any help please?

Hi,

I got this code which is meant to display the login details of the person that is logged in, however it just displays the details of the last person in the mysql table.

I have set up some test logins, so if I login as paul1 the details for paul3 are displayed...confused

Anyway, here is the page which displays the details

Code: [Select]
<?php

session_start();

// This checks to make sure the session variable is registered

// WARNING: DO NOT DELETE THE FOLLOWING LINE OF TEXT

if( isset($_SESSION['username']) && isset($_SESSION['sid'])) {

// You are free to edit the following code to suit your requirements
include_once("../../data/server.php");

include_once("../../lib/userdata.php");

// THIS BIT WORKS AND DISPLAYS THE USERNAME

 $data = mysql_query("SELECT * FROM members") 
 or die(mysql_error()); 
 while($info = mysql_fetch_array( $data )) 
 { 
 


include("../../lib/userinfo.php"); 
 
 //////////////////////////////////////////// WARNING: YOU SHOULD NOT EDIT ANYTHING ABOVE THIS LINE ////////////////////////////////////////////////////////
 
 
ECHO <<<PAGE

<!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>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>$siteName</title>
<link rel="stylesheet" href="../../userstylesheet.css" type="text/css" />

</head>

<div id="page">
  <img alt="" src="../../images/leftCurve" height="6" width="6" id="left" /> <img alt="" src="../../images/rightCurve.gif" height="6" width="6" id="right" />
  
  <div id="pageName">
    <h1>$siteName</h1>
     </div>
  <div id="pageNav">
    <div id="sectionLinks"> <a href="profile.php?username=$username">My Profile</a> <a href="modify.php?username=$username">Personal Details</a> <a href="message.php?username=$username">Messages</a> <a href="../../logout.php?username=$username">Logout</a></div>
  </div>
  <div id="content">
    <div class="feature">

<h2>Welcome $username </h2>
  

      <p>This is the demonstration home.html template. You are free to edit this or any of the other templates to suit your own needs. </p>
      <p>This is the first page your member will see once they have logged in. </p>
  <p>If you look at the code for this page, you will see that all HTML code is placed between the ***PAGE and PAGE; tags. Please note that the three * should be replaced with the < character. 
  This format must be kept to ensure that the user variables work. Changing this format may result in errors being returned.</p>
  <p>You may call member information using the $ tag and the variable name eg $ firstname without the space, will show the members first name, such as $firstname</p>
      <p>For any information please visit our site http://www.membersitemaker.co.uk. User guides will be added shortly and the forum will soon be full of help. </p>
    </div>
  </div>
  <div id="information">  <a href="#">About Us</a> | <a href="#">Site Map</a> | <a href="#">Privacy Policy</a> | <a href="#">Contact Us</a> | &copy;2011 $siteName </div>
</div>

</body>
</html>
PAGE;

}
//////////////////////////////////////// WARNING: DO NOT DELETE ANYTHING BELOW THIS LINE  //////////////////////////////////////////////////////////

} else {

// This will redirect the user to the login page if the session variables do not exist

header( "Location: ../../../login.html" );
}
?>
And here is the code for userdata.php

Code: [Select]
<?php


// Decode sitename

function decode_variable(&$siteName)
{
    $siteName = urldecode($siteName);
    $siteName = str_replace('%20',' ',$siteName); 
    return $siteName;
} 

decode_variable($siteName);

// Connnect to MySQL database

include_once("../../data/mysql.php");

$mysqlPassword = (base64_decode($mysqlpword));

$db = mysql_connect("$localhost", "$mysqlusername", "$mysqlPassword") or die ("Error connecting to database");

mysql_select_db("$dbname", $db) or die ("An error occured when connecting to database");

// Carry out MySQL query


?>
and userinfo.php

Code: [Select]
<?php

$username = $info['username'];
 $firstname = $info['firstname'];
 $lastname = $info['lastname'];
 $address = $info['address'];
 $town = $info['town'];
 $county = $info['county'];
 $postcode = $info['postcode'];
 $email = $info['email'];



$birth_year = $info['birth_year'];

$country = $info['country'];

$telephone_number = $info['telephone_number'];

$mobile_number = $info['mobile_number'];

$nickname = $info['nickname'];


As always, your help is much appreciated

Paul