Next_prev error

[Anon]

<?php

    if (!$newsstart) {
    $limit = "7";
    $nstart = "0";
    $nprev = "0";
    $nnext = "$limit";
    } else {
    $newprev = ($newsstart - $limit);
    $newnext = ($newsstart + $limit);
    }

    $con = mysql_connect("localhost", "$username", "$pass") or

die(mysql_error());
$db = mysql_select_db("pinehead",$con) or die(mysql_error());
$news = mysql_fetch_row(mysql_query("SELECT * FROM articles order by
artid desc $newsstart,$limit"));

    $con = mysql_connect("localhost", "$pass", "$username") or

die(mysql_error());
$db = mysql_select_db("pinehead",$con) or die(mysql_error());
$sql = "SELECT * FROM articles";
$result = mysql_query($sql,$con);
while ($art = mysql_fetch_array($result)) {
$title = $art['title'];
$content = $art['content'];

    echo $title;
    echo ("<br>");

}

echo ("<a href=\"?newsstart=$nprev\">Back</a> <a
href=\"?newsstart=$nnext\">Next</a>");

?>

Thats my code

Warning: Supplied argument is not a valid MySQL result resource in nav.php on line 17

thats the error

    $news = mysql_fetch_row(mysql_query("SELECT * FROM articles order by

artid desc $newsstart,$limit"));
thats line 17

Thanks for your help.
Anthony

[[deleted]]

You forgot the LIMIT clause in the query.

Use error checking:
http://www.xs4all.nl/~vinny/index.html#4

[Anon]

It looks to me like you forgot the resource link identifier. Isn't that represented by $con in your code?