[RESOLVED] corrupted result!?

[phpgoal]

function getid(){

$conn = db_connect();

$query = "...";
$fresult = mysql_query($query);

$count = mysql_numrows($fresult);

while($count!=0) {
	$num = mysql_fetch_array($fresult);
	if($num){
		$post['xnum'] = $num['xnum'];
		echo "$post[xnum],";	// RESULT = 62,64,65, BUT
	}
	$count--;
}
$post['xnum'] = $count;
return $post;

}

but in the calling function!?
function calltheotherf(){

$id = getid();
$count = $nums['count'];

for ($i = $count; $i>0; $i--) {
	$xnumid = $id['xnum'];
	echo " --[what #: $xnumid ]  ";
}

}
I get this!

--[what #: 65 ] --[what #: 6 ] --[what #: 6 ]

Anybody would know what the heck is going on?
Appreciate your feedback.

[Weedpacket]

Could you write that a bit more carefully (take all the time you need, this isn't instant messaging). Using [php]...[/code] tags around your code would be good for a start; that way it will keep its formatting and not look like the mess it looks like now (unless it really IS a mess, of course 😉).

I can't run your code, but even if I could, the second function wouldn't produce the results you claim because $nums is not defined. So I'm guessing that that is not the real code.

[phpgoal]

My apologies, it is not mess, it is just without the tags, sorry.
Here with the tags:

<?php
function getid(){		

$conn = db_connect();

$query = "...";
$fresult = mysql_query($query);

$count = mysql_numrows($fresult);

while($count!=0) {
	$num = mysql_fetch_array($fresult);
	if($num){
		$post['xnum'] = $num['xnum'];
		echo "$post[xnum],";	// RESULT = 62,64,65, BUT
	}
	$count--;
}
$post['xnum'] = $count;
return $post;
}


// but in the calling function!?
function calltheotherf(){

$id = getid();
$count = $id['count'];

for ($i = $count; $i>0; $i--) {
	$xnumid = $id['xnum'];
	echo " --[what #: $xnumid ]  ";
}
}

//I get this!

// --[what #: 65 ] --[what #: 6 ] --[what #: 6 ] 

// Anybody would know what the heck is going on?
// Appreciate your feedback.
?>