Question

~~最简单的~~什么是最有效的方式来获得下一个生日日期我有出生日期，如＆＃39; 1983-08-15＆＃39;

目前，我使用strtotime()

执行此类操作

function get_next_date($start_timestamp, $interval = 1, $time_frame = 'year'){

                                //+1 year
    $nextdate = strtotime('+'.$interval.' '.$time_frame, $start_timestamp);

       //date is still in the past
    if($nextdate - time() < 0){
         return get_next_date($nextdate);
    }

    return $nextdate;
}

效率不高（如果日期过去很多，则递归过多）。

我希望有一个解决方案，我可以轻松更改$interval和$time_frame

编辑：

建议的解决方案strtotime(date('d-M-', $start_timestamp).date('Y')." +{$interval} {$time_frame}")无效：

//(assuming today is the 2014-07-22)
1983-03-01 => 2015-03-01  //OK
1983-08-01 => 2015-08-01  //FALSE, should be 2014-08-01

例如，该函数也应该接受不同的间隔来获得每10个生日：

1983-03-01 => 2023-03-01
1984-08-01 => 2014-08-01

Answer 1

使用DateTime类的另一种方法：

function get_next_birthday($birthday) {
    $date = new DateTime($birthday);
    $date->modify('+' . date('Y') - $date->format('Y') . ' years');
    if($date < new DateTime()) {
        $date->modify('+1 year');
    }

    return $date->format('Y-m-d');
}

echo get_next_birthday('1983-08-15');

Answer 2

感谢@JonathanKuhn，我已经完成了这个功能：

function get_next_date($starttime, $interval, $time_frame) {

    $now = time();

    //based on the timeframe get the amount since the $startdate
    switch ($time_frame) {
        case 'year':
            $count = date('Y', $now)-date('Y', $starttime);
            break;
        case 'month':
            $count = abs((date('Y', $now) - date('Y', $starttime))*12 + (date('m', $now) - date('m', $starttime)));
            break;
        case 'week':
            $count = floor((abs($now - $starttime)/86400)/7);
            break;
        case 'day':
            $count = floor(abs($now - $starttime)/86400);
            break;

        default:
            //if you have other time frames you should add them here
            $count = $interval;
            break;
    }

    //how often the interval should get multipled
    $times = ceil($count/$interval);

    //get the next date by counting from the starting date up until $now with the calculated interval
    $nextdate = strtotime(date('d-M-Y', $starttime)." +".($interval*$times)." {$time_frame}");

    //date is maybe in the past (but in the same year, month or week)
    if($nextdate - $now < 0){
        //add an additional interval to the expression
        $nextdate = strtotime(date('d-M-Y', $starttime)." +".($interval*$times+$interval)." {$time_frame}");
    }

    return $nextdate;

}

从PHP的已知日期开始下一个即将到来的生日

2 个答案: