[SOLVED] PHP/MySQL Resources

[bdmovies]

Ridiculously stupid question, I know - but I just cant' quite wrap my head around this one.

 

I need to know how many open jobs an employee has, simple enough.

 

<?php
$sql = "SELECT COUNT(*) FROM cases WHERE open = '1' AND initiatedBy = '$_SESSION[employeeID]'";
$result = mysql_query($sql);
$openCases = mysql_num_rows($result);
echo $openCases;
?>

 

However, this returns 1 (the correct answer is 9...)

 

Why does echoing $result give me a Resource ID #? Shouldn't echoing $result show me what the query returned?

 

I'm using CocoaMySQL as well, and when I do a custom query using $sql It returns fine

 

I've also tried mysql_affected_rows.

 

What am I not seeing here?

 

[Maq]

Count(*) only returns how many records, not a record.  Can you please show me the fields in your database?  Is there perhaps a field similar to "open_jobs"?

[bdmovies]

I got it figured out. I read this topic:

 

http://www.phpfreaks.com/forums/index.php?topic=210087.0

 

<?php
$sql = 'SELECT COUNT(*) as OpenJobs FROM `cases` WHERE initiatedBy = \'1\' AND open = \'1\'';
$result = mysql_query($sql) or die (mysql_error());
$row = mysql_fetch_assoc($result);
$openCases =  $row['OpenJobs']; 
?>