display array in single result function question

[bluemonster]

Problem display record, no value found. if i use for each, can display, must i use for each to display?

$categoryContent = getCategoryContent($catId);
$catname=$categoryContent["catname"];

function getCategoryContent($catId = 0)
{
	$sql = "SELECT cat_id, cat_name, cat_desc, cat_files FROM categories WHERE cat_id='$catId'";
	$result =& dbQuery($sql) or die('Cannot get Record. ' . mysql_error());
$cat = array();
	$row = mysql_fetch_array($result);
	if ($row > 0) {

	if ($row["cat_files"]) {
		$cat_files = 'upload/category/' . $cat_files;
	} else {
		$cat_files = '';
	}
   $cat[0] = array(); 
   $cat[0]['catname'] = $row['cat_name']; 
   $cat[0]['catdesc'] = $row['cat_desc']; 
   $cat[0]['catfiles'] = $row['cat_files']; 
	//$cat[] = array($catname, $catdesc, $catfiles); 
	//$cat[] = array('url'   => $_SERVER['PHP_SELF'] . '?c=' . $catId,
	//			   'catname'  => $row["cat_name"],
	//			   'catdesc'  => $row["cat_desc"],
	//			   'catfiles'  => $row["cat_files"]);
}	

return $cat;
}

[reddrum]

I think you need somthing like this:

while ($Row = mysql_fetch_array ($result)) {
   $var[$i] = $Row['myField'];
   ++$i
}

[bluemonster]

reddrum wrote:

I think you need somthing like this:

while ($Row = mysql_fetch_array ($result)) {
   $var[$i] = $Row['myField'];
   ++$i
}

i think yours won't work, i just need to return 1 record. No need while?

[reddrum]

try:
$row = mysql_fetch_assoc($result);
$var[1] = $row['field1'];
$var[2] = $row['field2'];
ect.

[bluemonster]

can anyone answer? i mean it is working if only i use for each then it can display.

I try to use $catname=$categoryContent["catname"]; it wont display result.