php fetch array

[chriscloyd]

heres my error im getting

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\Program Files\Apache Software Foundation\Apache2.2\htdocs\includes\liveprojects.php on line 12

heres the code i have

[code]<?php
$get_num_project = mysql_query("SELECT * FROM projects WHERE status = 'Completed'");
$num_project = mysql_num_rows($get_num_project);
if ($num_project <= 5) {
$maxlimit = 5;
$minlimit = 1;
} else {
$maxlimit = rand(6,$num_project);
$minlimit = ($maxlimit - 5);
}
$project_get = mysql_query("SELECT * FROM projects WHERE status = 'Completed' LIMIT '$minlimit','$maxlimit'");
while ($p = mysql_fetch_array($project_get)) {
echo '<table width="95%" border="0" cellspacing="0" cellpadding="0">
  <tr>
    <td width="85%">'.$p['title'].'</td>
    <td align="center" width="15%"><a href=protfolio.php?project='.$p['id'].'">View</a></td>
  </tr>
</table>';
}
?>[/code]

[chronister]

[code]$project_get = mysql_query("SELECT * FROM projects WHERE status = 'Completed' LIMIT '$minlimit','$maxlimit'") or die(mysql_error());
[/code]
try that and see if any errors are occuring. I added the or die..... part.

[chriscloyd]

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''1','2'' at line 1

[chronister]

try this.
[code]
$project_get = mysql_query("SELECT * FROM projects WHERE status = 'Completed' LIMIT $minlimit,$maxlimit") or die(mysql_error());[/code]

No quotes needed around the $minlimit, $maxlimit part

[tauchai83]

it could be the query problem if u got that kind of msg...

[Orio]

You already opened a thread with this issue. I asked you try something but you never replied...
Read what I told you.
http://www.phpfreaks.com/forums/index.php/topic,124308.msg514803.html

Orio.