Re: making FORM dissapear when successful login

JHollis <jhollis@xxxxxxxxx> · Tue, 04 Jan 2005 23:13:42 -0500

Thank you for your suggestion, but do you care to be more descriptive, 
maybe some examples would be nice.  Can you tell my why the code i have 
doesnt work?  I would love to use sessions, but i was just trying to get 
an understanding of how the basics work first.

Jonathan wrote:
I think you should use session control after checking for login.

"JHollis" <jhollis@xxxxxxxxx> wrote in message
news:20050105020517.31324.qmail@xxxxxxxxxxxxxxx

I had this code working the way i wanted it to (as far as correct
username and password allowing successful login)...but what i want to
happen now is when a user successfully logs it it will make the login
form disappear and just say successfully logged in or welcome user and a
link below it so they can log off and make the form re-appear.  Below is
the code that i have where i tried to get it to disappear on successful
login, but it stays disappeared all the time.  Can someone please point
out what im doing wrong.  I have tried everything i can think of...and
nothing works.  Im a PHP newbie...so im sure some of you might get a
laugh out of this...if it is real easy.

<---snippet

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd";>
<html>
<head>
<link href="style.css" rel="stylesheet" type="text/css" />
</head>
<body>
 <div id="container">
<div id="top">
<h1>header</h1>

</div>
<div id="leftnav">
<p>

<?php

$username=$_POST['username'];
$password=$_POST['password'];
$db="user";
$server="localhost";
$db_username="root";
$db_password="*******";

$connect = mysql_connect($server,$db_username,$db_password);
if (!$connect) {
die ("could not connect to database");
}
$select = mysql_select_db($db,$connect);
if (!$select) {
die ("could not select database $db");
}
 /*username='$username'";*/
$sql = "SELECT * FROM passwords, user_info where id=PID and
username='$username'";
$result = mysql_query($sql);
/*$num_rows = mysql_num_rows($result);*/
while ($user = mysql_fetch_array($result))
{
$id = $user['id'];
$username2 = $user['username'];
$password2 = $user['password'];
$firstname = $user['firstname'];
$email = $user['email_address'];

IF ($username=="$username2" && $password=="$password2")
{
echo("\"Welcome, <b>$firstname</b>\"");?><br><?
echo ("\"Your email address is <b>$email<b>\"");?></td><tr>
<a href="<?$_SERVER['PHP_SELF']?>?username=""&?password=""">Logoff</a><?
break;
}
else
{
?>
<FORM action="<?$_SERVER['PHP_SELF']?>" method="post">
    <INPUT type="hidden" name="id">
<table>
<td><b>*</b>Username:</td> <td><INPUT class="input" size="8"
type="text" name="username" value="<?echo $username?>"></td><tr>
<td><b>*</b>Password:</td> <td><INPUT class="input" size="8"
type="password" name="password"></td><tr>
<td class="xsmall"><b>* Case Sensitive</b></td>
<td><INPUT type="submit" value="Login"></td><tr>
<td>&nbsp </td>
</table>
</FORM>
<?

break;
}

}
//IF ($username != $username2 || $password != $password2) {//

?><br>
<?
if  ($username == "" && $password == "") {
echo ("Please type in a Username and Password");}

if  ($username != "" && $password == "") {
echo ("Please type in a password");}

if  ($username == "" && $password != "") {
echo ("Please type in a username and password");}

?>

</p>
</div>
<?if (($username2==$username && $password2==$password) &&
($username2!="" || $password2!="")){?>
<div id="rightnav" class="box">
<p>
</p>
</div>
<?}?>
<div id="content">
<h2>Subheading</h2>
<p>
</p>
<p>
</p>
</div>
<div id="footer">
<p>
Today is
<?php
echo( date("F dS Y."));
?>
</p>
</div>
</div>
</body>
</html>
---->snippet

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