Antal links ud for kategorier

$res = mysql_query("SELECT COUNT(*) AS antal, lcat_name FROM links, category WHERE links.cat_id=category.sub_cat AND links.link_val='YES' GROUP BY links.cat_id");
while ($row = mysql_fetch_array($res)) {
echo "Name: $row[1] - Antal: $row[0]<br>\n";
}

Er det sådan noget?

Sådan ser det ud i øjeblikket men jeg kan ikke få den til at virke helt som forventet? Solution

$rime = mysql_query("SELECT * from category WHERE sub_cat='0' ORDER BY lcat_name")
or die(mysql_error());

echo "<tr>";
$i = 1;
while($row = mysql_fetch_array($rime)) {
extract($row);


$id = $row["lcat_id"];
$name = $row["lcat_name"];

$navigator = ".." ;

$rime1= mysql_query("SELECT * FROM category WHERE sub_cat='$id' LIMIT $nosubcat");
while($row = mysql_fetch_array($rime1)) {
extract($row);
$ida = $row["lcat_id"];
$namea = $row["lcat_name"];
$sub_cat = $row["sub_cat"];
$linnk = "<a href='$PHP_SELF?cat=$lcat_id'>$namea</a>";
$navigator = $linnk ." ".$navigator;

}

//$rime11 = mysql_query("SELECT COUNT(*) FROM links, category WHERE '$id'=category.lcat_id AND links.link_val = 'YES'") or die(mysql_error());
//$totalnum1 = mysql_result($rime11, 0, 0);

$rime11 = mysql_query("SELECT COUNT(*)AS antal, lcat_name FROM links, category WHERE '$id'=category.lcat_id AND links.link_val='YES' GROUP BY links.cat_id") or die(mysql_error());
while($totalnum1 = mysql_fetch_array($rime11)){

echo "<td width='$cinuz' valign=top><a href='links.php?cat=$id'><font class='text10'><b>$name</b></font></a>$totalnum1[0]<br>";

echo "<font class='text7'>$navigator</font><br><br /></td>";
}

har selv fundet løsningen

tog sin tid lol