Help W/ Pagination Code From Library

[ShoeLace1291]

I need help with this pagination code that I got from the PHP Code Library at PHPFreaks.com. 

 

This is my image moderation section of the admin panel i'm making:

if($manage == 'images'){

$date = date('M j, o H:i:s');
$time = strtotime($date);
$datetime=date('Y-m-d',$time);

           $session = mysql_query("UPDATE img_mod_sessions SET datetime='$datetime' WHERE uid='$uid'") or die("Error: ".mysql_error());


echo "<h1>Image Moderation</h1>";

$sql = mysql_query("SELECT filename,userID,sid FROM recent_files ORDER BY sid DESC LIMIT 10") or die("Pagination: ".mysql_error());

include("functions/pagination.php");

    $pageid        = $_GET['pid'];//Get the pid value

    if(intval($pageid) == 0) $pageid  = 1; 

    $Paging        = new paging();

    $total_records = $Paging->myRecordCount($sql);//count records

    $totalpage     = $Paging->processPaging(12,$pageid);

    $result        = $Paging->startPaging($sql);//get records in the databse

    $links         = $Paging->pageLinks(basename($PHP_SELF));//1234 links

    unset($Paging); 

$numimgs = mysql_num_rows($sql);

if($numimgs == 0){

     echo "No new images have been uploaded since your last image moderation session.";

} else {

     while($fetch=mysql_fetch_array($query)){

          $filename=$fetch["filename"];
          $posterid=$fetch["posterID"];

     $query2 = mysql_query("SELECT username FROM users WHERE uid = '$posterid'") or die("Error: ".mysql_error());

            $u=mysql_fetch_array($query2);

               $postername=$u["username"];

         echo "<a href='image.php?img=$filename'><img src='uploads/$filename' alt='$filename' width='50' height='50' border='0'></a><br>Posted by $postername<br><br>";

     }

  }

}

 

This is functions/pagination.php:

<? class paging
{
var $pRecordCount;
var $pStartFile;
var $pRowsPerPage;
var $pRecord;
var $pCounter;
var $pPageID;
var $pShowLinkNotice;
function processPaging($rowsPerPage,$pageID)
{
	       $record = $this->pRecordCount;
		   if($record >=$rowsPerPage)
		    	$record=ceil($record/$rowsPerPage);
		   else
				$record=1;
			if(empty($pageID) or $pageID==1)
			{
				$pageID=1;
				$startFile=0;
			}
			if($pageID>1)
			$startFile=($pageID-1)*$rowsPerPage;
			$this->pStartFile   = $startFile;
			$this->pRowsPerPage = $rowsPerPage;
			$this->pRecord      = $record;
			$this->pPageID      = $pageID;
			return $record;
}
function myRecordCount($query)
{
	$rs      			= mysql_query($query) or die(mysql_error()."<br>".$query);
	$rsCount 			= mysql_num_rows($rs);
	$this->pRecordCount = $rsCount;
	unset($rs);
	return $rsCount;
}
function startPaging($query)
{
	$query    = $query." LIMIT ".$this->pStartFile.",".$this->pRowsPerPage;
	$rs       = mysql_query($query) or die(mysql_error()."<br>".$query);
	return $rs;
}
function pageLinks($url)
{
	$this->pShowLinkNotice = " ";
	if($this->pRecordCount>$this->pRowsPerPage)
	{		
		$this->pShowLinkNotice = "Page ".$this->pPageID. " of ".$this->pRecord;
		//Previous link
		if($this->pPageID!==1)
		{
		$prevPage = $this->pPageID - 1;
		$link = "<a href=\"$url?$pid=1&mid=$ltype&cid=$catid\" class=\"$cssclass\">|<<</a>";
		$link .= "<a href=\"$url?$pid=$prevPage&mid=$ltype&cid=$catid\" class=\"$cssclass\"><<</a>     ";
		}
		//Number links 1.2.3.4.5.
		for($ctr=1;$ctr<=$this->pRecord;$ctr++)
		{	
			if($this->pPageID==$ctr)
			$link .=  " <a href=\"$url&pid=$ctr\" style=\"font-family:Verdana, Arial, Helvetica, sans-serif; font-size:10px; color:#F5E000\"><b>$ctr</b></a>";
			else
			$link .= " <a href=\"$url&pid=$ctr\" style=\"font-family:Verdana, Arial, Helvetica, sans-serif; font-size:10px; color:#F5E000\">$ctr</a>";
		}
		//Previous Next link
		if($this->pPageID<($ctr-1))
		{
		$nextPage = $this->pPageID + 1;
		$link .= "     <a href=\"$url?$pid=$nextPage&mid=$ltype&cid=$catid\" class=\"$cssclass\">>></a>";	
		$link .="<a href=\"$url?$pid=".$this->pRecord."&mid=$ltype&cid=$catid\" class=\"$cssclass\">>>|</a>";
		}
		return $link;
	}			
}

}
?>

 

I am getting a MySQL Syntax error that says 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 'Resource id #10' at line 1

Resource id #10.  I am guessing it is in pagination.php since only that file uses that type of error output(with the resource id).  I have no idea why it's doing this.  Thanks for any help in advance.

[chigley]

$total_records = $Paging->myRecordCount($sql);//count records

 

You're sending an already completed query to be re-run, the OOP will run it for you.

 

Change:

 

$sql = mysql_query("SELECT filename,userID,sid FROM recent_files ORDER BY sid DESC LIMIT 10") or die("Pagination: ".mysql_error());

 

To:

 

$sql = "SELECT filename,userID,sid FROM recent_files ORDER BY sid DESC LIMIT 10")

[ShoeLace1291]

Ok, now I'm getting this error: 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 'LIMIT 0,12' at line 1

SELECT filename,userID,sid FROM recent_files ORDER BY sid DESC LIMIT 10 LIMIT 0,12

[chigley]

If you look at your query, first you say "select the first ten rows" then straight after you say "select the first twelve rows"! You can only limit by one value: so either LIMIT 10 or LIMIT 12.

[ShoeLace1291]

Ok, I'm confused.  What do I need to change now?

[chigley]

Well it depends if you want to select the 10 most recent files or 12 most recent files!

[ShoeLace1291]

I want to display the 12 most recent files.

[chigley]

$sql = "SELECT filename,userID,sid FROM recent_files ORDER BY sid DESC LIMIT 12";