将天数转换为年，月，日格式

Question

I have this problem. Can someone help me,how to convert number of days into the format XX Years, XX Month, XX Days... i created this function,

function convert($sum) {
    $years = ($sum / 365) ;
    $years = floor($years); 
    $month = ($sum % 365) / 30.5; 
    $month = floor($month); 
    $days = ($sum % 365) % 30.5; // the rest of days
    // Echo all information set
    echo 'DAYS RECEIVE : '.$sum.' days<br>';
    echo $years.' years - '.$month.' month - '.$days.' days';
}

convert(151);

But with 151 days the result was wrong

DAYS RECEIVE : 151 days
0 years - 4 month - 1 days

it must be 4 month ans 28 days not 1 day...

http://sandbox.onlinephpfunctions.com/code/f5e6b4b4f6a27024b66ffbf04e80698722a3ecab

Answer 1

If you use more modern PHP, the following is based around actual days in each month:

$days = 151;

$start_date = new DateTime();
$end_date = (new $start_date)->add(new DateInterval("P{$days}D") );
$dd = date_diff($start_date,$end_date);
echo $dd->y." years ".$dd->m." months ".$dd->d." days";

Note that it will vary, depending on the current date, so you might prefer to set $start_date and $end_date to work from a fixed baseline

$days = 151;

$start_date = new DateTime('1970-01-01');
$end_date = (new DateTime('1970-01-01'))->add(new DateInterval("P{$days}D") );
$dd = date_diff($start_date,$end_date);
echo $dd->y." years ".$dd->m." months ".$dd->d." days";

Answer 2

This is the solution for your problem:

function convert($sum) {
    $years = floor($sum / 365);
    $months = floor(($sum - ($years * 365))/30.5);
    $days = ($sum - ($years * 365) - ($months * 30.5));
    echo “Days received: ” . $sum . “ days <br />”;
    echo $years . “ years, “ . $months . “months, “ . $days . “days”;
}