PHP：生日檢查今天的日期

Question

我有用戶的生日保存在birthday爲1999-02-26。

如何檢查生日是否是今天？

if(date('m-d') == ..? 

Answer 1

if(date('m-d') == substr($birthday,5,5)) 

要添加什麼蒂姆說：

if(date('m-d') == substr($birthday,5,5) or (date('y')%4 <> 0 and substr($birthday,5,5)=='02-29' and date('m-d')=='02-28')) 

Answer 2

這個答案應該的工作，但是這取決於strtotime能找出你的數據庫的日期格式：

$birthDate = '1999-02-26'; // Read this from the DB instead 
$time = strtotime($birthDate); 
if(date('m-d') == date('m-d', $time)) { 
    // They're the same! 
} 

Answer 3

從PHP 5.2起：

if (substr($dateFromDb, -5) === date_create()->format('m-d')) { 
    // Happy birthday! 
} 

Answer 4

<?php 
/** 
* @param string $birthday Y-m-d 
* @param int $now 
* @return bool 
*/ 
function birthdayToday($birthday, $now = null) { 
    $birthday = substr($birthday, -5); 
    if ($now === null) { 
     $now = time(); 
    } 
    $today = date('m-d', $now); 
    return $birthday == $today || $birthday == '02-29' && $today == '02-28' && !checkdate(2, 29, date('Y', $now)); 
}