Question

我有age = 39，month = 10和day = 5，我需要获得出生年份

这一年需要考虑age，month和day

任何想法？

Answer 1

你可以试试这个：

<?php
  $age = 21;
  $birthmonth = 8;
  $birthday = 26;
  $thisyear = date('Y', time());
  $checkdate = "$thisyear-$birthmonth-$birthday";
  if((time()-strtotime($checkdate)) > 0){
       $birthyear = $thisyear - $age;
  } else {
       $birthyear = $thisyear - $age - 1;
  }
  $fullstring = "$birthyear-$birthmonth-$birthday";
  $fullstring = date('m/d/Y', strtotime($fullstring));
  echo $fullstring;
?>

它有点笨重，但它会起作用。它检查他们的生日是否已经发生并计算出来。

Answer 2

我刚刚用我的DOB创建了一个小例子，它可以正常工作，希望这就是你要找的东西

$age = 34 ;
$mon = 01;
$day = 29 ;

$beck_year = date("Y",strtotime("- $age years"));

$dob = $beck_year.'-'.$mon.'-'.$day ;

Answer 3

在评论中扩展纠正我错误的逻辑......

if( $month >= date("m") && $day >= date("d")
    $year = date("Y") - $age;
else
    $year = $date("Y") - $age - 1;

Answer 4

检查一下希望这会有所帮助！

$years = 10;
$months = 1;
$days = 1;

function get_byear($years,$months,$days ){
    $b_year = date('Y', strtotime("$years years $months months $days days ago"));
    return $b_year;
}

echo get_byear($years,$months,$days);

如何在PHP中获得年龄，月份和日期的年份？

4 个答案: