Question

我的数据库包含'birthdate'列，用户可以用“Y-m-d”格式填充。

现在我想要回复希望用户生日快乐的总天数。

E.g。如果当前日期是2014-06-06并且用户出生日期是1980-06-26，那么生日即将到来6月26日，代码应该回应“生日是在20天后”

我试过的是跟随 -

$date1= date("m-d");
$date2=date("m-d",strtotime($data['birthdate']));

$dateDiff = abs(strtotime($date2) - strtotime($date1));
$remainedDays = floor($dateDiff/(60*60*24));

echo "Birthday Is After $remainedDays Days";

但是这段代码给了0天

Answer 1

执行此操作的最简单方法是DateTime classes，特别是DateTime::diff method。请注意，您应该确保每个日期的时间是午夜，然后测试今年的生日是否已经发生。

<?php
$birthday = "1980-06-24";

// get date of birthday this calendar year
$parts = explode('-', $birthday, 2);
$birth_date = new DateTime(date('Y') . '-' . $parts[1] .' 00:00:00');
$today = new DateTime('midnight today');

if ($birth_date < $today) {
    // next birthday is in one year
    $birth_date->modify("+1 Year"); 
}

// get number of days days remaining
$diff = $birth_date->diff($today);

if ($diff->days > 0) {
    echo "There are " . $diff->days . " remaining until your birthday.";
} else {
    echo "Happy birthday!";
}

Answer 2

这是一个可用于即将到来的B'days的算法。这也将考虑闰年，B'day为02/29

您可以根据需要修改代码。它将检查已经过期的即将到来的B'day或B'day。

$bday  = new DateTime("1980-06-26");
$today  = new DateTime();
$b_y = $bday->format('Y');
$b_d = $bday->format('d');
$b_m = $bday->format('m');

if((bool)$bday->format('L') && $b_d == 29 && $b_m == 2){
  if((bool)$today->format('L')){
    $bday_obj = new DateTime(date('Y').'-'.$b_m.'-'.$b_d);
    $diff = $bday_obj->diff($today);
  }else{
    for($i=1;$i++;$i<=3){
      $today->add(new DateInterval('P1Y'));
      if((bool)$today->format('L')){
        $bday_obj = new DateTime(date('Y').'-'.$b_m.'-'.$b_d);
        $diff = $bday_obj->diff($today);
        break ;
      }
    }
  }
}else{
  $bday_obj = new DateTime(date('Y').'-'.$b_m.'-'.$b_d);
  $diff = $bday_obj->diff($today);
}

$now = new DateTime() ;
if($now > $bday_obj){
  echo 'Your birthday was over '.$diff->format('%a days').' before';
}else{
  echo 'Your birthday is after  '.$diff->format('%a days');
}

计算出生日和当天的天差

2 个答案: