Rewrite this:

$result = mysql_query($query);

Like this:

if (!$result = mysql_query($query)) {
    print "Could not run query ($query) : " . mysql_error();
    exit;
}

If that doesn't output something, try and rewrite your query like so:

$query = "SELECT count(*) 
          FROM books 
          WHERE $searchtype
          LIKE '%$searchterm%'";

And get the count by:

if ($result = mysql_query($query)) {
    print mysql_result($result,0);
    exit;
}

Anyway this looks like a support question but just in case try the above.  I'm guessing the query is invalid, like, $searchtype is not defined correctly or something.  In which case you'd be giving mysql_num_rows() a invalid mysql result resource because mysql_query() retured false on the bogus query.  And BTW, you want && not ||.  

Error handling is your friend.  Like for example, make sure $searchtype is a column you want to use.  Anyway, when in doubt, print stuff.

Okay, I did the first rewrite and I got the below output.
 Could not run query (select * from books where Java 2 like '%Java 2%') : You have an error in your SQL syntax near '2 like '%Java 2%'' at line 1

Then I tried the second rewrite and I got the one below.
Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /bookorama/results.php on line 32

You can't have spaces in column names, your query is bogus.  And btw I just noticed your typo:

$searchtype = addslashes($searchtype);
$searchtype = addslashes($searchterm);  // notice the problem?

Anyway, this is bogus.

This is what my code looks like now:
<?
	trim($searchterm);
	if(!$searchtype || !$searchterm) {
		echo "You have not entered search details. Please go back and try again.";
		exit;
	}
	
	$searchtype = addslashes($searchtype);
	$searchterm = addslashes($searchterm);
	
	@ $db = mysql_connect("localhost", "headdive_jazz", "jazz");
	
	if(!$db) {
		echo "Error: Could not connect to database. Please try again later.";
		exit;
	}
	
	mysql_select_db("books");
	$query = "SELECT count(*) FROM books WHERE $searchtype LIKE '%$searchterm%'";
				
	if(!$result = mysql_query($query)) {
		print mysql_result($result, 0);
		exit;
	}
	
	$num_result = mysql_num_rows($result);
	
	echo "<p>Number of books found: ".$num_results."</p>";
	
	for($i=0; $i < $num_results; $i++) {
		$row = mysql_fetch_array($result);
		echo "<p><strong>".($i + 1).". Title: ";
		echo htmlspecialchars(stripslashes($row["title"]));
		echo "</strong><br>Author: ";
		echo htmlspecialchars(stripslashes($row["author"]));
		echo "<br>ISBN: ";
		echo htmlspecialchars(stripslashes($row["isbn"]));
		echo "<br>Price: ";
		echo htmlspecialchars(stripslashes($row["price"]));
		echo "</p>";
	}
?>

and this is the response I get:
Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /bookorama/results.php on line 29

Please ignore everything I said and go here:

http://www.php.net/support

This is not a bug in PHP.