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year函数 有关问题,在线给分

WBOY
WBOYOriginal
2016-06-13 13:53:41894browse

year函数 问题,在线给分

如$year=过的年数,如5
$staff_join=开始的年月日 如2008-05-02
我现要 把开始的年月日加上年数,如 2008-05-02+5=2013-05-02 用什么表达式呀,
要可运行的代码,谢谢

小弟这样 ($Begin=(date("Y",$staff_join)+$year) 出错, 输出1970)

------解决方案--------------------

PHP code

<?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);
?>
<br><font color="#e78608">------解决方案--------------------</font><br>这里赋值写错了<br>$staff_join ="20010-05-02";<br>改成$staff_join ="2010-05-02";
<br><font color="#e78608">------解决方案--------------------</font><br>
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 $arrdate[0]; //2013
?>
<br><font color="#e78608">------解决方案--------------------</font><br><pre class="brush:php;toolbar:false"><br><?php <br />$year = 5;<br>$staff_join='2008-05-02';<br><br>$join_date = mktime(0, 0, 0,  <br>	substr($staff_join, 5, 2), substr($staff_join, -2, 2), substr($staff_join, 0, 4));<br>echo date('Y-m-d', $join_date), "\n";<br>$join_date_5_years_later =  <br>	strtotime("+$year Year", $join_date );<br><br>echo date('Y-m-d', $join_date_5_years_later);<br>?><br>

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

<?php $value = 5;
$year = date('Y');
$year += $value;
echo $year.date('-m-d');
<br /><font color="#e78608">------解决方案--------------------</font><br>$year = 5;  <br>$staff_join = '2008-05-02';<br><br>$Begin = date("Y-m-d", strtotime("+$year year $staff_join));<br>
<br><font color="#e78608">------解决方案--------------------</font><br>给楼上老大改下<br><?php   <br />$year = 5;  <br>$staff_join = '2008-05-02';  <br><br>$Begin = date("Y-m-d", strtotime("+$year year $staff_join"));  <br>echo $Begin;<br>?>
<br><font color="#e78608">------解决方案--------------------</font><br><?php   <br />$year = 5;  <br>$staff_join = '2008-05-02';  <br><br>$Begin = date("Y-m-d", strtotime("+$year year,$staff_join"));  <br>echo $Begin;<br>?>
<br><font color="#e78608">------解决方案--------------------</font><br>看来已经解决了
<br><font color="#e78608">------解决方案--------------------</font><br>strtotime
<br><font color="#e78608">------解决方案--------------------</font><br>老兄去研究一下strtotime函数,手册上有详细说明。 <div class="clear">
                 
              
              
        
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