>when I echo menu($id) I get the current page's title.
>
>How do I print it's peers and it's single parent?
You will need a second query to ask for all the children of the parent:
$peer_query = "select id as child_id, title from meta_data where pid =
$pid";
$peers = mysql_query($query) or error_log(mysql_error());
while (list($child, $child_title) = mysql_fetch_row($peers)){
echo "$child_title ($child)<BR>\n";
}
>P.P.P.S. All pages have information provided by this script:
>
>$fn = explode("/", $_SERVER['PHP_SELF']);
> $num_of_s = count($fn) - 1;
> $fn = "$fn[$num_of_s]";
> $query = "SELECT * FROM meta_data WHERE page_name = '$fn'";
> $result = mysql_query($query);
> $num_results = mysql_num_rows($result);
> $row = mysql_fetch_array($result);
> $id = $row['id'];
> $pid = $row['pid'];
> $title = $row['title'];
> $description = $row['description'];
> $keywords = $row['keywords'];
>
>It is my metadata page and is used all over the place. so why re-query the
>db for it's parent?
I'm saying:
Don't do that. Don't use * in this first query. In *ANY* query, specify
*exactly* which columns you need. You'll be a lot less confused by your
data when you start forcing yourself to be more precise in your code about
what you want.
*CHANGE* the line above to:
$query = "select id, pid, title, description, keywords from meta_data where
page_name = '$fn'";
Then, after you have the $pid, add in the stuff I wrote above.
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