Ver Mensaje Individual
  #1 (permalink)  
Antiguo 30/04/2009, 05:29
rtr
 
Fecha de Ingreso: abril-2009
Mensajes: 368
Antigüedad: 15 años
Puntos: 2
el profile.php siempre me da lo mismo....

En el profile.php siempre me da los mismos datos , aunque cree varias cuentas y entre con diferentes nombres.....siempre me da los mismos datos....¿sabe alguien porque?
Deberia verse los datos de cada cuenta...¿no?
Os dejo el código..

[PHP]<?php
$username = $_COOKIE['loggedin'];
if (!isset($_COOKIE['loggedin'])) die("You are not logged in, <a href=../login.html>click here</a> to login.");
echo "You are logged in $username";
?> | <a href="index.php">Return to members home</a>
<p>
<br />
<form action="update.php" method="post" >
<p>Username:
<input type="text" name="username" disabled="disabled" value="<?php
include "../config.php";
mysql_connect($server, $db_user, $db_pass) or die (mysql_error());
$result = mysql_db_query($database, "select * from $table WHERE username = '$username'") or die (mysql_error());

while ($qry = mysql_fetch_array($result)) {
echo "$qry[username]";
}
?>" />
</p>
<p>Change your password</p>
<p> Old Password:
<input type="password" name="oldpass" />
<br />
New Password:
<input type="password" name="newpass" />
<br />
</p>
Your Personal Details
<p>First Name:
<input type="text" name="name1" value="<?php
include "../config.php";
mysql_connect($server, $db_user, $db_pass) or die (mysql_error());
$result = mysql_db_query($database, "select * from $table WHERE username = '$username'") or die (mysql_error());

while ($qry = mysql_fetch_array($result)) {
echo "$qry[name1]";
}
?>" />
</p>
<p>Second Name:
<input type="text" name="name2" value="<?php
include "../config.php";
mysql_connect($server, $db_user, $db_pass) or die (mysql_error());
$result = mysql_db_query($database, "select * from $table WHERE username = '$username'") or die (mysql_error());

while ($qry = mysql_fetch_array($result)) {
echo "$qry[name2]";
}
?>" />
</p>
<p>Email Address:
<input type="text" name="email" value="<?php
include "../config.php";
mysql_connect($server, $db_user, $db_pass) or die (mysql_error