แก้รูปแบบวันที่จาก d/m/y เป็น y-m-d ต้องทำอย่างไรบ้างคะ

date('Y-m-d', strtotime("30/10/2011"));

function convertdate($s_date,$s_from,$s_to,$s_return_delimiter) {
	$s_return_date = '';
    $s_from = strtolower($s_from);
    $s_to = strtolower($s_to);
    $s_date = str_replace(array('\'', '-', '.', ',', ' '), '/', $s_date);
    $a_date = explode('/', $s_date); 
    
    switch($s_from) {
         case 'eng': # dd/mm/yyyy
             $day = $a_date[0];
             $month = $a_date[1];
             $year = $a_date[2]; 
        break;
         case 'usa':  # mm/dd/yyyy
             $month = $a_date[0];
             $day = $a_date[1];
             $year = $a_date[2]; 
        break;
         case 'iso': # yyyy/mm/dd
             $year = $a_date[0]; 
            $month = $a_date[1];
             $day = $a_date[2];
         break;
         default: # error message
             user_error('function convertdate(string $s_date, string $s_from, string $s_to, string $s_return_delimiter) $s_from not a valid type of \'eng\', \'usa\' or \'iso\'');
             return NULL;
     }
 
    # substitution fixes of valid alternative human input e.g. 1/12/08
     if (strlen($day)==1) { $day='0'.$day; } # day -trailing zero missing
     if (strlen($month)==1) { $month='0'.$month; } # month -trailing zero missing
     if (strlen($year)==3) { $year=substr(date('Y'),0,strlen(date('Y'))-3).$year; } # year -millennium missing
     if (strlen($year)==2) { $year=substr(date('Y'),0,strlen(date('Y'))-2).$year; } # year -century missing
     if (strlen($year)==1) { $year=substr(date('Y'),0,strlen(date('Y'))-1).$year; } # year -decade missing
 
    switch($s_to) {
         case 'eng': # dd/mm/yyyy
             $s_return_date = $day.$s_return_delimiter.$month.$s_return_delimiter.$year;
         break;
         case 'usa':  # mm/dd/yyyy
             $s_return_date = $month.$s_return_delimiter.$day.$s_return_delimiter.$year;
         break;
         case "iso": # yyyy/mm/dd
             $s_return_date = $year.$s_return_delimiter.$month.$s_return_delimiter.$day;
         break;
         default: # error message
             user_error('function convertdate(string $s_date, string $s_from, string $s_to, string $s_return_delimiter) $s_to not a valid type of \'eng\', \'usa\' or \'iso\'');
             return NULL;
     }
 
    # if it's an invalid calendar date e.g. 40/02/2009 or rt/we/garbage
     if (!is_numeric($month) || !is_numeric($day) || !is_numeric($year)) { 
        return NULL;
     } elseif (!checkdate($month, $day, $year)) {
         return NULL;
     }
 
    return $s_return_date;
 } 

//echo convertdate('13/04/2009','eng','iso','-');

$date_reg = '31/12/2011';
list($dd,$mm,$yy)=explode("/",$date_reg);
				if(checkdate($mm,$dd,$yy))
				{
					$Date = convertdate($date_reg,'eng','iso','-');
                                        echo "$Date";
                                }