year函数有关问题，在线给分

------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

";
echo $arrdate[0]; //2013
?>

------解决方案--------------------

$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);
?>


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

------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
$year = 5;  
$staff_join = '2008-05-02';  

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

------解决方案--------------------
$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。