如何编写整洁的代码来获取上一个生日日期？

Question

I know the day of birth, lets say: 1996-12-12
I know that today is: 2014-11-12

Now I want to get the date of the last birthday. And I have written this solution, but I don't like the complexity of the code. How can I write it more readable for other programmers?

$lstrBirthday       = '1996-12-12';
$lstrLastBirtday    = (strtotime(date('Y').date('-m-d', strtotime($lstrBirthday))) > strtotime('now')) ? (date('Y') - 1).date('-m-d', strtotime($lstrBirthday)) : date('Y').date('-m-d', strtotime($lstrBirthday));

Thanks!

Answer 1

Its easier if you use the DateTime functions

$begin = new DateTime( '1996-12-12' );
$end = new DateTime();
$end = $end->modify( '-1 day' ); 

$interval = new DateInterval('P1Y');
$daterange = new DatePeriod($begin, $interval ,$end);

$bd = '';

foreach($daterange as $date){
    $bd = $date->format("Ymd");
}

Edit: So now you have the last entry ;)

Something like that i can't test it now but you can foreach over the period. If the date is in the past you should get the last birthday. If you make +1 year its in the future.

Edit: For @pascalvgemert perhaps its a bit more readable for you...

public function getLastBirthdate($startDate) {
    $begin = new DateTime( $startDate );
    $end = new DateTime();
    $end = $end->modify( '-1 day' ); 

    $interval = new DateInterval('P1Y');
    $daterange = new DatePeriod($begin, $interval ,$end);

    $bd = '';

    foreach($daterange as $date){
        $bd = $date->format("Ymd");
    }

    return $bd;
}

$lastBD = getLastBirthdate('1996-12-12');