Hi, I'm new to this forum (but not really to PHP). I'm not sure if my problem is a variable problem or what, but here's the code I have:

<? 
if (isset($type)) {
$link = mysql_connect($host, $user, $pass) or die("Could not connect.");
mysql_select_db($db, $link) or die("Couldn't select database.");

$query = "SELECT * from $issue WHERE type='$type'";

$result = mysql_query($query, $link);

while ($row = mysql_fetch_array($result)) {
echo "<font size='3'><b>".$row['title']."</b></font><br>";
echo "<p align='right'>Written By: ".$row['author']."</p>";
echo $row['content'];
echo "<p>";
}
} else {
echo "Sorry, but you stumbled into the wrong area.";
}
?>

I keep recieving an error: "Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in c:\program files\apache group\apache\htdocs\yoraritard\main.html on line 18" (Line 18 is the while ($row = mysql_fetch_array($result)).)

The variable $type is passed through the browser (the address would be main.html?type=whatever).

Any suggestions?

I don't see where $issue is set.

You will need to get the 'type' from the browser -

$type = $_GET['type'];

Secondly, I also do not see where you have set $issue as a variable. If that is not set, the query does not know what table to look at.

Steve.

hudster, both Phaedrus and Gaffer point out good points -- $issue and $type are never set. Use print to debug your sql and also make the following call to see how many rows you are getting back:

PHP Code:

print "<p>Number of results returned by SQL query:  " . mysql_num_rows( $result ); 


However, if $result is not a valid mysql resource (that is, if the query failed), this call will produce the same error.

I would display what $query looks like, perhaps the variables are not being set as expected or maybe there is something in there (quotes?) that may be throwing the query off

PHP Code:

$query = "SELECT * from $issue WHERE type='$type'";
print "sql = " . $query;