year函数 有关问题，在线给分

<?php
$year = 5;
$staff_join ="20010-05-02";
$a = strtotime("+".$year." Year")-time(); 
$staff_join = strtotime($staff_join)+$a;
echo date('Y-m-d',$staff_join);
?>

------解决方案--------------------
这里赋值写错了
$staff_join ="20010-05-02";
改成$staff_join ="2010-05-02";

------解决方案--------------------
PHP code

<?
$year = 5;
$staff_join = '2008-05-02';
$arrdate= explode('-',$staff_join);
$arrdate[0] += $year;
$staff_out = implode('-',$arrdate);
echo $staff_out; //2013-05-02
echo "<br>";
echo $arrdate[0]; //2013
?>

------解决方案--------------------
<pre>
<?php
$year = 5;
$staff_join='2008-05-02';

$join_date = mktime(0, 0, 0,  
	substr($staff_join, 5, 2), substr($staff_join, -2, 2), substr($staff_join, 0, 4));
echo date('Y-m-d', $join_date), "\n";
$join_date_5_years_later =  
	strtotime("+$year Year", $join_date );

echo date('Y-m-d', $join_date_5_years_later);
?>
</pre>

------解决方案--------------------
PHP code

<?php 
$value = 5;
$year = date('Y');
$year += $value;
echo $year.date('-m-d');

------解决方案--------------------
$year = 5;  
$staff_join = '2008-05-02';

$Begin = date("Y-m-d", strtotime("+$year year $staff_join));


------解决方案--------------------
给楼上老大改下
<?php  
$year = 5;  
$staff_join = '2008-05-02';  

$Begin = date("Y-m-d", strtotime("+$year year $staff_join"));  
echo $Begin;
?>

------解决方案--------------------
<?php  
$year = 5;  
$staff_join = '2008-05-02';  

$Begin = date("Y-m-d", strtotime("+$year year,$staff_join"));  
echo $Begin;
?>

------解决方案--------------------
看来已经解决了

------解决方案--------------------
strtotime

------解决方案--------------------
老兄去研究一下strtotime函数，手册上有详细说明。