[SOLVED] a mysql_fetch error using php

[chriscloyd]

heres my error

[quote]Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/srxstud/public_html/test22/includes/liveprojects.php on line 7
[/quote]
heres my code

[code]<?php
$get_num_project = mysql_query("SELECT * FROM projects WHERE status = 'Completed'");
$num_project = mysql_num_rows($get_num_project);
$maxlimit = rand(1,$num_project);
$minlimit = $maxlimit - 5;
$project_get = mysql_query("SELECT * FROM projects WHERE status = 'Completed' LIMIT '$minlimit','$maxlimit'");
while ($p = mysql_fetch_assoc($project_get)) {
echo '<table width="95%" border="0" cellspacing="0" cellpadding="0">
  <tr>
    <td width="85%">'.$p['title'].'</td>
    <td align="center" width="15%"><a href=portfolio.php?project='.$p['id'].'">View</a></td>
  </tr>
</table>';
}
?>[/code]

[Orio]

It may be that $minlimit is negative and that's why there's an error.
Try:
[code]$minlimit = (($maxlimit - 5) > 0) ? $maxlimit - 5 : 0;[/code]

Orio.

[chriscloyd]

i tried that it didnt work

[Jessica]

$project_get = mysql_query("SELECT * FROM projects WHERE status = 'Completed' LIMIT '$minlimit','$maxlimit'") OR die(mysql_error());