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)
[解决办法]
<?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);?>
[解决办法]
这里赋值写错了
$staff_join ="20010-05-02";
改成$staff_join ="2010-05-02";
[解决办法]
<?$year = 5;$staff_join = '2008-05-02';$arrdate= explode('-',$staff_join);$arrdate[0] += $year;$staff_out = implode('-',$arrdate);echo $staff_out; //2013-05-02echo "<br>";echo $arrdate[0]; //2013?>
[解决办法]
<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>
[解决办法]
<?php $value = 5;$year = date('Y');$year += $value;echo $year.date('-m-d');
[解决办法]
$year = 5;
$staff_join = '2008-05-02';
$Begin = date("Y-m-d", strtotime("+$year year $staff_join));
[解决办法]
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;
?>
[解决办法]
<?php
$year = 5;
$staff_join = '2008-05-02';
$Begin = date("Y-m-d", strtotime("+$year year,$staff_join"));
echo $Begin;
?>
[解决办法]
看来已经解决了
[解决办法]
strtotime
[解决办法]
老兄去研究一下strtotime函数,手册上有详细说明。