kan ikke finde fejlen i dette script

hej eksperter...

Jeg har siddet og gloet mig blind på dette script. jeg kan simpelthen ikke finde ud af hvor min fejl er. så har brug for et par friske øjne til at kigge det igennem.

min fejl meddelelse kommer her.
Parse error: syntax error, unexpected '>' in /customers/d/0/9/jp-pro.dk/httpd.www/puchstoana/inc/billeder.php on line 7

siden hvor fejlen skal findes er denne.

billeder.php

<table border="0" cellpadding="0" cellspacing="0" width="700" align="center">
<?php
if($_GET['id]) {
$headline = mysql_query("select * from puch_gallari where gallari_id="'.$_GET['id'].'"") or die(mysql_error());
if(mysql_num_rows($headline)) {
while($hil = mysql_fetch_array($headline)) {
echo '<tr><td align="center" height="30"><strong>'.$hil['title'].'</strong></td></tr>';
}
}
}
?>
<?php
// how many rows to show per page
$rowsPerPage = 13;

// by default we show first page
$pageNum = 1;

// if $_GET['page'] defined, use it as page number
if(isset($_GET['page']))
{
    $pageNum = $_GET['page'];
}

// counting the offset
$offset = ($pageNum - 1) * $rowsPerPage;

$query = " SELECT * FROM puch_pictures where gallari='".$_GET['id']."'" .
        " LIMIT $offset, $rowsPerPage";
$result = mysql_query($query) or die('Error, query failed');

// print the random numbers
while($eve = mysql_fetch_array($result))
{
?>
<?php
$antal_cat = 3;
$count_cat = 0;
$query_cat = mysql_query("SELECT * FROM puch_pictures order by picture_id") or die(mysql_error());
if(mysql_num_rows($query_cat)>0)
{
    //Hoved Tabel Start
    echo '<table width="700"><tr>'."\r\n";
    while ($pic = mysql_fetch_array($query_cat))
    {
        if(($count_cat % $antal_cat)==0)
        {
            echo '</tr><tr>'."\r\n";
        }
     
        //Kategori Tabel Start
        echo '<td valign="top"><table border="0" cellpadding="0" cellspacing="0" width="100" background="#" height="100">'."\r\n";

            echo '<tr>

                  <td width="100" height="100"><a href="index.php?sidevalg=se_billed&pic_id='.$pic['picture_id'].'"><img src="img/upload/'.$pic['picture'].'" width="100" height="100"></a></td>

              </tr></table>'."\r\n";

$count_cat++;
    }
 
    //Hoved Tabel Stop
    echo ''."\r\n";
}
?>


<?php
}
$query  = "SELECT COUNT(*) AS numrows FROM puch_pictures";
$result  = mysql_query($query) or die('Error, query failed');
$row    = mysql_fetch_array($result, MYSQL_ASSOC);
$numrows = $row['numrows'];

// how many pages we have when using paging?
$maxPage = ceil($numrows/$rowsPerPage);

// print the link to access each page
$self = $_SERVER['PHP_SELF'];
$nav  = '';

for($page = 1; $page <= $maxPage; $page++)
{
  if ($page == $pageNum)
  {
      $nav .= " $page "; // no need to create a link to current page
  }
  else
  {
      $nav .= " <a href=\"$self?sidevalg=se_gallari&page=$page\">$page</a> ";
  }
}

if ($pageNum > 1)
{
  $page  = $pageNum - 1;
  $prev  = " <a href=\"$self?sidevalg=se_gallari&page=$page\">[Forrige]</a> ";

  $first = " <a href=\"$self?sidevalg=se_gallari&page=1\">[F&oslash;rste side]</a> ";
}
else
{
  $prev  = '&nbsp;'; // we're on page one, don't print previous link
  $first = '&nbsp;'; // nor the first page link
}

if ($pageNum < $maxPage)
{
  $page = $pageNum + 1;
  $next = " <a href=\"$self?sidevalg=se_gallari&page=$page\">[N&aelig;ste]</a> ";

  $last = " <a href=\"$self?sidevalg=se_gallari&page=$maxPage\">[Sidste side]</a> ";
}
else
{
  $next = '&nbsp;'; // we're on the last page, don't print next link
  $last = '&nbsp;'; // nor the last page link
}

echo $first . $prev .
" Viser side $pageNum af $maxPage sider " . $next . $last;
// ... and we're done!
?>

Håber i kan finde fejlen for jeg kan ikke.

På forhånd tak for hjælpen

Venlig hislen
Delphiuser