如何在php中獲得年齡，月份和日期？

Question

我有age = 39，month = 10和day = 5，我需要得到出生年份

年需要採取在考慮age，month和day

什麼想法？

Answer 1

你可以試試這個：

<?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; 
?> 

這是一個有點笨拙，但它會工作。它會檢查他們的生日是否已經發生並計算出來。

Answer 2

我剛剛創建了一個小例子，我的出生日期和它的作品，希望這是你在找什麼

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

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

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

Answer 3

擴展在修正我的意見有缺陷的邏輯......

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

Answer 4

檢查這一項，希望這將有助於！

$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);