Data Not Selecting?

[koomann]

<?php
include("accesscontrol.php" ); 
include("common.php" );
include("db.php" );
dbConnect("star-craftx_com" );
$sql = "SELECT FROM design_users WHERE
username='$username' AND password='$password' AND aim='$aim' AND email='$email' date_joined='$date_joined'";

$date_joined = date("M d, Y h:i A", "$date_joined" );
echo "Welcome<b> $username</b>.<br>";
echo "You registered on:<b> $date_joined </b><br>";
echo "<b>Contact Info:</b><blockquote><br>";
echo "<b>AIM:</b> $aim <br>";
echo "<b>ICQ:</b> $icq <br>";
echo "<a href="mailto:$email">Email <b> $username </b></blockquote>";

?>

This is what it prints out:

Welcome koomann.
You registered on: Dec 31, 1969 06:00 PM
Contact Info:

AIM:
ICQ:
Email koomann

I don't know what I'm going wrong...

(NOTE: the reason why the $username still shows up, is because of "accesscontrol.php", the included file at the top).

[PHP John]

I'm afraid I don't see the problem. Could you point out just what is not working?

[koomann]

It dosen't pring out the information: $icq, $aim, $email ect. ect.

www.star-craftx.com/practice/xdesigns/bob.php

login is: koomann/blah
The date is also out of wack... >=\

[PHP John]

I would bet that for some reason these variables are not being assigned a value.

You will have to dig into those includes to find out where they were supposed to be assigned.

[koomann]

But the includes don't assign any variables, except for "accesscontrol.php", which carries over the $username

db.php - dbConnect function...
common.php - hold an error function for when a part of a form isn't filled out, it'll prompt up a small message box, then bring them back to the form.

[JeffWalden]

I don't understand then, where are the $aim and $icq variables supposed to be coming from? There is no source for them.

Maybe I'm just being ignorant.

[PHP John]

Exactly. Just where are these variables being assigned?

[koomann]

Doesn't this do it?

$sql = "SELECT * FROM design_users WHERE
date_joined='$date_joined' AND icq='$icq' AND aim='$aim' AND email='$email'"; 

[PHP John]

Doesn't this do it?

$sql = "SELECT * FROM design_users WHERE
date_joined='$date_joined' AND icq='$icq' AND aim='$aim' AND email='$email'"; 

No. $date_joined, $icq, and $aim all need to have a value assigned to them in order for that query to even work.

What you want to do is to retrieve that information FROM the database.

$result      = mysql_query( "SELECT * FROM design_users WHERE user_name = '$user_name'" );
$resultArray = mysql_fetch_array( $result );
$date_joined = $resultArray['date_joined'];
$icq         = $resultArray['icq'];
$aim         = $resultArray['aim']; 

Now you can use the $date_joined, $icq, and $aim variables because they now have a value to return.