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 $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>
$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>
<?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"> </div>