Sætte celler ved siden af hinanden

$n =1;
echo "<table>\n<tr>\n";
while($row = mysql_fetch_array($res)) {
  if ($n == 2) {
    echo "</tr>\n<tr>\n";
  }
  echo "<td>".$row."</td>\n";
  $n++
}
echo "</table>\n</tr>\n";

Hov.. min fejl.. Her (+ at hvis det er et ulige antal celler bliver den valid):

$n = 1;
echo "<table>\n<tr>\n";
while($row = mysql_fetch_array($res)) {
  echo "<td>".$row."</td>\n";
  if ($n == 2) {
    echo "</tr>\n<tr>\n";
    $n = 1;
  }
  $n++
}
if ($n == 2) {
  echo "<td></td>";
}
echo "</table>\n</tr>\n";

Hmm den virker ikke helt
den siger

Parse error: parse error, unexpected '}' in c:\www\pc-dk.dk\index.php on line 14

<table border="1">
<?php
include("dvd/cfg/dbconnect.php");
$res = mysql_query("SELECT * FROM dvd") or die(mysql_error());
$n = 1;
echo "<table>\n<tr>\n";
while($row = mysql_fetch_array($res)) {
    echo "<td>".$row."</td>\n";
    if ($n == 2) {
        echo "</tr>\n<tr>\n";
        $n = 1;
    }
    $n++
}

if ($n == 2) {
  echo "<td></td>";
}
echo "</table>\n</tr>\n";
?> 

Linie 13 er
den } efter $n = 1;

havde lige glemt ; efter $n++ ;-)
<table border="1">
<?php
include("dvd/cfg/dbconnect.php");
$res = mysql_query("SELECT * FROM dvd") or die(mysql_error());
$n = 1;
echo "<table>\n<tr>\n";
while($row = mysql_fetch_array($res)) {
    echo "<td>".$row."</td>\n";
    if ($n == 2) {
        echo "</tr>\n<tr>\n";
        $n = 1;
    }
    $n++;
}

if ($n == 2) {
  echo "<td></td>";
}
echo "</table>\n</tr>\n";
?>

Den er der næsten
:-P
http://www.pc-dk.dk

(Burde måske teste dem før jeg sender dem ;-) )
Men her..

<table border="1">
<?php
include("dvd/cfg/dbconnect.php");
$res = mysql_query("SELECT * FROM dvd") or die(mysql_error());
$n = 1;
echo "<table>\n<tr>\n";
while($row = mysql_fetch_array($res)) {
  echo "<td>".$row."</td>\n";
  if ($n == 2) {
      echo "</tr>\n<tr>\n";
      $n = 1;
  } else {
  $n++;
  }
}

if ($n == 2) {
  echo "<td></td>";
}
echo "</table>\n</tr>\n";
?>

takker for det her er dine point

selv tak ;-)