Displaying SQL Table

[VictorProfundus]

Hello!

I have been working on an affiliate link system. A user would click a php link that would add 1 to the count of a certain affiliate, by ID and then return the user to that affiliate page.

My problem comes when I wish to display a table for my affiliates for them to see how many clicks they have and such.

Right now, here is my code for the table :

<?php

  echo "<table border=1 cellpadding=0 cellspacing=3><tr><td align=center>Affiliate ID Number</td><td align=center>Name</td><td align=center>Number of Clicks</td><td align=center>Payment</td><td align=center>Comments</td></tr>";

  for ($i=0; $i<$num; $i++) {
	  if ($tmp = mysql_fetch_array($result)) {
		  extract($tmp);

		  echo "<tr><td align=center>$id</td>";
		  echo "<td align=center>$name</td>";
		  echo "<td align=center>$count</td>";
		  echo "<td align=center>$paid</td>";
		  echo "<td align=center>$comments</td></tr>";
	  }
  }

  echo "</table>";

?>

For some reason this isn't working with my page.

Does anyone have a simpler way of doing this?

[ronverdonk]

Whatever happened to the code leading up to the $result? Maybe the error is in your select, so it won't display anything. Show that code leading to the $result resource id.

Ronald

[VictorProfundus]

okay...here is something like what the entire page would look like:

<?php

include ('header.php');


  mysql_connect("localhost", "username", "P*A*S*S*W*O*R*D");
  @mysql_select_db("my database") or die( "Unable to connect to database");
  $result = mysql_query("SELECT * FROM linkcount");

  $num = mysql_num_rows($result);

  mysql_close();

include ('table.php');

?>

The table.php then would be my previous post.

[ak1dnar]

Hi pal, just put your mysql_close() function end of the page.
[code=php]
<?php
include ('table.php');
mysql_close();
?>
[/code]