Hello, I have this silly bug.
<code>
<?
function staffEcho($infoType){
$query = Mysql_Query(\"SELECT * FROM staff\", $connection);
$rows = Mysql_Num_Rows($query);
for ($a=0; $a<$rows; $a++)
{
$echoThis = Mysql_Result($query, $a, $infoType);
echo($echoThis . \"<br>\");
}
}
?>
</code>
which is called with
<code>
<? staffEcho(\"staff_nick_name\");?>
</code>
gives the error
\"Warning: Supplied argument is not a valid MySQL-Link resource in includes/staff.php on line 5\"
When everything is in one place and it is done like
<code>
<?
function staffEcho($infoType){
$query = Mysql_Query(\"SELECT * FROM staff\", $connection);
$rows = Mysql_Num_Rows($query);
for ($a=0; $a<$rows; $a++)
{
$echoThis = Mysql_Result($query, $a, \"staff_nick_name\");
echo($echoThis . \"<br>\");
}
}
?>
</code>
it works fine. It\'s when I turn it into a function that I call it doesn\'t work. Any ideas .. ?
Thanks in advance!
