Php / mysql dynamically generated multi user website

I need a php / mysql programmer to fix the errors in the functions.php page. You will find a copy of the functions.php below. The code is 90% correct. I am presently using it but there are errors.

The functions.php is still looking in too many places. Only fk_accounts 24, 25, and 26 hold that data is needed for the website.
25 = Name
24 = email
26 = ID #

The code is looking in too many places which is causing errors and omissions.

The functions.php retrieves each member's data by looking up the "id=" querrystring. Such as:
http://www.uc2.biz/isb.php?t=1&id=000-03-9425
Each member has their own personal dynamically generated website.

If you are willing to fix the functions.php code, I will give you a quick overview of what the code is to do, although that should be self evident.

Best Regards
Liberate



<?php

##functions.php

function dbc() {
//connect to mysql and choose database.
mysql_connect("mysql1.xxxxxxxxxx.com","xxxxxxx","xxxxxxx");
mysql_select_db("maillist");
}

dbc();

function checkid() {
global $id;
//is this ID# a valid one from the DB?
$result = mysql_query("select value from swd_doppar where value='$id'") or die(mysql_error());
if(mysql_num_rows($result) > 0) {
return 1;
}else{
return 0;
}
}

if(!checkid()) {
print("<script Language=\"JavaScript\">location.href=('http://www.uc2.biz/adm/not_registered.php');</script></NO SCRIPT>");

exit;
}

function getuid($id) {
global $uid;
//get fk_user value from db table 'swd_doppar' based on $id
$result = mysql_query("select value,fk_user from swd_doppar where value='$id'") or die(mysql_error());
$row = mysql_fetch_array($result);
$uid = $row['fk_user'];
return $uid;
}

dbc();
$uid = "";
getuid($id);

function printName() {
global $uid;
//locator id's
$id1 = 25;
$id2 = 17;
$id3 = 21;

//get name
$qry = "select * from swd_doppar where fk_user='$uid' and fk_fields='25'";
$result = mysql_query($qry) or die(mysql_error());
$result = mysql_query("select * from swd_doppar where fk_user='$uid' and fk_fields='$id1'");
if(mysql_num_rows($result) == 0) {
$result = mysql_query("select * from swd_doppar where fk_user='$uid' and fk_fields='$id2'");

if(mysql_num_rows($result) == 0) {
$result = mysql_query("select * from swd_doppar where fk_user='$uid' and fk_fields='$id3'");
}
}
$row = mysql_fetch_array($result);
print($row[2]);
}

function printEmail($toggle) {
global $uid;
//locator id's
$id1 = 24;
$id2 = 20;
$id3 = 24;

//get email
$result = mysql_query("select * from swd_doppar where fk_user='$uid' and fk_fields='$id1'");
if(mysql_num_rows($result) == 0) {
$result = mysql_query("select * from swd_doppar where fk_user='$uid' and fk_fields='$id2'");

if(mysql_num_rows($result) == 0) {
$result = mysql_query("select * from swd_doppar where fk_user='$uid' and fk_fields='$id3'");
}
}


$row = mysql_fetch_array($result);

//display email in a way that it won't be harvested...
//explode email into two different parts

if($toggle==1) {
$expld = explode("@",$row[2]);
?>
<SCRIPT language="JavaScript" type="text/javascript">
<!--
var name = "<? print($expld[0]); ?>";
var domain = "<? print($expld[1]); ?>";
document.write('<a href=\"mailto:' + name + '@' + domain + '\">');
document.write(name + '@' + domain + '<\/a>');
// -->
</SCRIPT>
<?
}else{
$rec_email = $row[2];
return $rec_email;
}
}
?>

is this a paid fix?
the link to my online resume:
tellmama.com:800/cds/resumeComp.html
Email : wndrer@earthling.net