PHP Forum

**jwgrafflin** · 09:55 PM

Here's the scenario. I have numerous affiliate programs I promote through affiliate networks such as Commission Junction, Cleverbridge and LinkShare. Some of those affiliate programs list with more than one affiliate network.

I list the individual program in my database as "progname" and the associated affiliate network(s) as "netname1", "netname2" and "netname3" with corresponding referral urls for the associated affiliate program as "netnameurl1", "netnameurl2" and "netnameurl3."

My desired output is a list of affiliate programs and their matching referral url where the affiliate network name matches the search criteria.

For example, let's say affiliate program "freelancer.com" is listed with both Commission Junction and LinkShare and my search term is Commission Junction. For that particular program, LinkShare is entered as "netname1" and Commission Junction is entered as "netname2" (network listing order may vary for other programs). My desired output would be:

freelancer.com http://www.cj.com/?id=jwg

If my search term was LinkShare, the desired output would be:

freelancer.com http://www.ls.com/?id=gra

Here's what I have so far, but it won't list the url.

Code:

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" xmlns:v="urn:schemas-microsoft-com:vml" >
<head>
<meta name="robots" content="noindex, nofollow"/>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8"/>
<link rel="stylesheet" type="text/css" href="http://www.jwgmg.com/styles/style2.css"/>
<title>Search Results for Affiliate Program Database</title>
<base target="_blank" />
</head> 
<body>
<div style="width: 100%; padding: 1em; text-align: center; margin: auto">
<p class="center bold lg em">Program by Network Search Results</p>
<?
include("dbinfo.inc.php");
 
$search="".$search."";
 
$query = "SELECT * FROM programs WHERE netname1 LIKE '$search' OR netname2 LIKE '$search' OR netname3 LIKE '$search' ORDER BY progname ASC";
$result=mysql_query($query)or die(mysql_error());
 
$num=mysql_num_rows($result);
$i=0;
 
 
?>
<h1><?echo $search;?></h1>
<table style="border: 1px solid black; text-align: center; margin: auto">
<thead style="background-color: #369; color: white">
<tr>
<th>Program Name</th>
<th>Referral URL</th>
</tr>
</thead> 
 
<?
while($i< $num){
$progname=mysql_result($result,$i,"progname");
if($netname1=="$search"){$neturl=mysql_result($result,$i,"netnameurl1");}
elseif($netname2=="$search"){$neturl=mysql_result($result,$i,"netnameurl2");}
elseif($netname3=="$search"){$neturl=mysql_result($result,$i,"netnameurl3");}
?>
 
<tbody class="left">
<tr>
<td><?echo $progname;?></td>
<td><?echo $neturl;?></td>
</tr>
 
<?
++$i;
}
?>
</tbody> 
</table>
<table style="margin-left: 12em; border: 2px solid red">
<tr><td style="background-color: #369; color: white"><?echo $num;?></td></tr>
</table>
</div> 
</body>
</html>

**PaulW** · 07-02-2011, 09:25 PM

"it won't list the url." Why not? Is it being mischievous? On strike, perhaps?

Seriously, not enough info. Are there correct values on the db to expect output of a certain form? You output $num - is that correct? What, if anything, is displayed in your generated table cells?

Given you're testing for equality change your if/elseifs to a swithc and put in a default branch to get a bit more diagnostic info.

**jwgrafflin** · 07-02-2011, 10:12 PM

Everything but the url is generated as desired.

**adityamenon90** · 07-05-2011, 01:06 PM

You might want to try this:

Code:

$query = "SELECT * FROM programs WHERE netname1 LIKE '%$search%' OR netname2 LIKE '%$search%' OR netname3 LIKE '%$search%' ORDER BY progname ASC";

Adding % symbol before and after your search terms makes it a "containing" search. Otherwise, your search is just the same as an "=" search. LIKE is looking for a regex pattern. So maybe enough results are not being generated? Maybe that's the problem?

**jwgrafflin** · 07-05-2011, 01:48 PM

No need for that. The correct output is there, it just won't display.

**jwgrafflin** · 08-27-2011, 11:03 PM

The "mischievous" code is no longer on "strike" and is behaving as it should once one minor little error was corrected. Problem solved, so you can close this thread.

**vectorialpx** · 08-28-2011, 02:49 AM

No harm feelings but, your coding is a mess.

Quote:

$search="".$search.""; // ???!!! what's this?!

Anyway... I can also see that your database structure is... same as coding. What if you will need netnameurl8 ?

My way is much more simple

Code:

 // You have the result
while($record = mysql_fetch_assoc($result)){
$progname = $record['progname'];
// the below condition is a real nonsense
// what's about $netname1 ? $netname2 ? $netname3 ?
// are they initialized somewhere
/*
if($netname1=="$search"){$neturl=mysql_result($result,$i,"netnameurl1");}
elseif($netname2=="$search"){$neturl=mysql_result($result,$i,"netnameurl2");}
elseif($netname3=="$search"){$neturl=mysql_result($result,$i,"netnameurl3");}
*/
$neturl = $record['netnameurl1'] .  $record['netnameurl2'] .  $record['netnameurl3'];
// it's not what you want but you can debug better
?>
 
<tbody class="left">
<tr>
<td><?echo $progname;?></td>
<td><?echo $neturl;?></td>
</tr>