Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/ickius/public_html/news.php on line 5

PHP Code:

<? 
mysql_connect("localhost","-","-");
mysql_select_db("-");

while($r=mysql_fetch_array(mysql_query("SELECT * FROM news ORDER BY id DESC"))) {
$title=$r["title"];
$date=$r["date"];
$name=$r["name"];
$story=$r["story"];

echo "<h1>$title</h1>\n";
echo "<p>$date posted by $name</p>\n";
echo "<p>$story</p>\n";
echo "<br />\n";
}
?>

What could be wrong? I'm so confused, it used to work fine.

BEcause you've wrapped up the mysql_query() call inside the while loop condition, it is creating a new query and a new result set every time round. Try this:

PHP Code:

<? 
mysql_connect("localhost","-","-");
mysql_select_db("-");

$result = mysql_query("SELECT * FROM news ORDER BY id DESC") or die(mysql_error());

while($r=mysql_fetch_array($result)) {
$title=$r["title"];
$date=$r["date"];
$name=$r["name"];
$story=$r["story"];

echo "<h1>$title</h1>\n";
echo "<p>$date posted by $name</p>\n";
echo "<p>$story</p>\n";
echo "<br />\n";
}
?>

This way you will step through the results one by one rather than always getting the first line of a new result set. The error message you are getting usually indicates that your query is failing for some reason. The call to mysql_error() should print the MySQL error and tell you why it is failing.

It says no database selected...? What's that mean, I've selected the database already.
Edit: Don't worry, I think it mite something with my host.