Resource id #3 HELP!!!!

This site is best viewed in a modern browser with JavaScript enabled.

g4ujs

Hi,

I am a total novice, trying to just return a count from a table, but I keep getting a Resource id #3, I see this is a common problem but I dont understand stuff I have read to know how to modify my code.

<?php
require('LogbookSearchConfig.php');
$db=MYSQL_CONNECT($hostname, $username, $password) OR DIE("Unable to connect to database");

@mysql_select_db("$dbName") OR DIE(error(mysql_error()));

    $call = isset($_POST['call']) ? $_POST['call'] : '';
    $MyCall = isset($_POST['MyCall']) ? $_POST['MyCall'] : '';
    $cate = isset($_POST['cate']) ? $_POST['cate'] : '';
                    if ($call == "")
                {$call = '';}
                    if ($MyCall == "")
                {$MyCall = '';}

$query = 'SELECT count(*) FROM `OnlineLogBook` WHERE date >= \'2013/01/01\''; 

        $result = @MYSQL_QUERY($query);
        $num_rows = @mysql_num_rows($result); 

        echo "$result";

          MYSQL_CLOSE($db);
                    exit;
   ?>

I know the SQL works but getting the number out into a webpage is defeating me!!!

Help!

regards

Bob

OctoberWind

It's showing "Resource #3" because $result is just that. a MySQL Resource.

http://www.php.net/manual/en/function.mysql-query.php

For SELECT, SHOW, DESCRIBE, EXPLAIN and other statements returning resultset, mysql_query() returns a resource on success, or FALSE on error.

The returned result resource should be passed to mysql_fetch_array(), and other functions for dealing with result tables, to access the returned data

You need to loop over the result with:

while ($row = mysql_fetch_assoc($result)) {
    echo $row["count(*)"];
}

Although, I'm not sure why you would need to do that, because your query would return the same thing as the mysql_num_rows.