[Resolved] pagination of search terms not working...why

[llitten]

I am posting a script that works for pagination.

I want to change it so that I can send THIS script search terms...such as specific columns in a table, specific search terms.

When I change the $query_count to something like...
$query_count = "SEARCH * FROM $usertable WHERE p_keywords LIKE %search_term%";

I get errors. I even tried the search_term in single quotes.

HOW CAN I ADD TO THIS PAGINATION SCRIPT SO THAT I CAN SEARCH OVER A SELECT COLUMN AND FIND SELECT INFORMATION????

Thanks for your time.
Lowell

	<p><?

include("files/picturedata.inc");

$page=$_GET["page"]; //my addition

@mysql_connect($hostname, $username, $password) or die("ERROR--CAN'T CONNECT TO SERVER");
@mysql_select_db($dbName) or die("ERROR--CAN'T CONNECT TO DB");

$limit          = 9;               
$query_count    = "SELECT * FROM $usertable";    //$query_count    = "SELECT count(*) FROM $usertable";    
$result_count   = mysql_query($query_count);    
$totalrows      = mysql_num_rows($result_count);

if(empty($page)){
    $page = 1;
}


$limitvalue = $page * $limit - ($limit);
$query  = "SELECT * FROM $usertable LIMIT $limitvalue, $limit";        
$result = mysql_query($query) or die("Error: " . mysql_error());

if(mysql_num_rows($result) == 0){
    echo("Nothing to Display!");
}

$bgcolor = "#E0E0E0"; // light gray


echo("<table width='100%' border ='0' cellspacing ='5' cellpadding='5' align='center'>");

$x=0;    //for  page format

while($row = mysql_fetch_array($result)){

    if ($x ==3) {
      print ("<tr>");         
      $x=0;
      }

print ("<td>");
print ("<div align = 'center'>");
print ("<a href='singlephoto.html?photoid=$row[0]'><img width='188' src='../photopages/photos/$row[0].jpg'></a>");
print ("<br>");
print ("$row[1]"); print ("<br>");

print ("<img width ='66' src='images/spacer.gif'>");
print ("</div>");
print ("<td>");

$x++;

}
echo("</table>");

if($page != 1){
    $pageprev = $page-1;
  //  $pageprev = $pageprev --;

   echo("<a href=\"$PHP_SELF?page=$pageprev\">PREV".$limit."</a>&nbsp;");
}else{
    echo("PREV ".$limit."&nbsp;");

}

$numofpages = $totalrows / $limit;

for($i = 1; $i <= $numofpages; $i++){
    if($i == $page){
        echo($i."&nbsp;");
    }else{
        echo("<a href=\"$PHP_SELF?page=$i\">$i</a>&nbsp;");          
    }
}


if(($totalrows % $limit) != 0){
    if($i == $page){
        echo($i."&nbsp;");
    }else{
        echo("<a href=\"$PHP_SELF?page=$i\">$i</a>&nbsp;");
    }
}

if(($totalrows - ($limit * $page)) > 0){
    $pagenext = $page+1;

    echo("<a href=\"$PHP_SELF?page=$pagenext\">NEXT".$limit."</a>");
}else{
    echo("NEXT".$limit);
}

mysql_free_result($result);

?></p>
<p></p>
</body>

</html>

[llitten]

the exact message states:

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result

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 'LIKE %nikki%' at line 1