I'm writing a simple weblog, and I'm new to PHP; but thanks to this particular resource, I'm learning more and more. With that said, I'm making some mistakes, which I need help finding solutions for. I'm receiving these three Warning/Errors:

Warning: mysql_query() [function.mysql-query]: Access denied for user 'root'@'localhost' (using password: NO) in /Users/laurenyoung/Sites/blogtastic/viewcat.php on line 20

Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in /Users/laurenyoung/Sites/blogtastic/viewcat.php on line 20

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /Users/laurenyoung/Sites/blogtastic/viewcat.php on line 22

And my code is:

<?php
require("includes/config.php");

if(isset($_GET['id']) == TRUE) {
   if(is_numeric($id) == FALSE) {
	$error = 1;
}
if($error == 1) {
   header("Location: " . $config_basedir . "/viewcat.php");
}
else {
   $validcat = $_GET['id'];
}
	}
else {
   $validcat = 0;
}

$sql = "SELECT * FROM categories";
$result = mysql_query($sql);

while($row = mysql_fetch_assoc($result)) {
   if($validcat == $row['id']) {
	echo "<strong>" . $row['cat'] . "</strong><br />";		
	$entriessql = "SELECT * FROM entries WHERE cat_id = " . $validcat ." ORDER BY dateposted DESC;";
	$entriesres = mysql_query($entriessql);
	$numrows_entries = mysql_num_rows($entriesres);
		
	echo "<ul>";
		
	if($numrows_entries == 0) {
		echo "<li>No entries!</li>";
		}
		else {
			while($entriesrow = mysql_fetch_assoc($entriesres)) {				echo "<li>" . date("D jS F Y g.iA", strtotime($entriesrow['dateposted'])) ." - <a href='viewentry.php?id=" . $entriesrow['id'] . "'>" . $entriesrow['subject'] ."</a></li>";
						}
		}
		echo "</ul>";
   }
   else {
	echo "<a href='viewcat.php?id=" . $row['id'] . "'>" . $row['cat'] .
		"</a><br />";
	}
}
require("includes/footer.php");
?>

Could someone <kindly> explain what my issue is here?

Thanks

Check for connection string in your script. Its unable to connect. $connection =mysql_connect("host","user","password") or die (mysql_error()); Will give you the correct error message.

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