Confusing mySQL issue?

[abcorn]

So, I created an account here to ask this question.

 

I have this code:

<?php
function userlist() {
echo '<select name="user" id="user">';
$q = mysql_query("SELECT * FROM `users` ORDER BY last_name ASC, first_name ASC");
while($a = mysql_fetch_array($q)) {
	echo '<option value="'.$a['id'].'">';
	echo $a['first_name'].' '.$a['last_name'];
	echo '</option>';
}
echo '</select>';
}
?>

 

The table `users` looks like this:

• id - int(11)
  
• email - varchar(50)
  
• first_name - varchar(50)
  
• last_name - varchar(50)
  
• phone - varchar(20)
  
• adminpassword - varchar(35)
  

 

The idea is, whenever I need a dropdown of all the users in the database, I can just call that function from within the form. For example,

 

<form method="post" action="foo.php">
<?php userlist(); ?>
<input type="submit" value="Submit" />
</form>

 

However, whenever I do this, it gives me a select with no options. A view-source reveals this:

<select name="user" id="user"><br /> 
<b>Warning</b>:  mysql_fetch_array(): supplied argument is not a valid MySQL result resource in <b><<file>></b> on line <b><<line>></b><br /> 
</select>

 

 

I am almost certain I am doing this right. I have used this kind of code many times before. I even copypasta'd the contents of mysql_query() into phpMyAdmin, and it returned the list fine. So what am I doing wrong?

 

Thanks in advance!

[requinix]

Right after the mysql_query() put a

echo mysql_error();

and see what error message you get. Could it be that you're missing the bit that connects to the database?

[abcorn]

Aha! Thanks. I'm building on someone else's server, and they have a calendar script that runs and apparently connects to its own database after I do mine. So I'll just reconnect.