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)+$year) 出错, 输出1970) 解决方案 » 免费领取超大流量手机卡,每月29元包185G流量+100分钟通话, 中国电信官方发货 <?php$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);?> 这里赋值写错了$staff_join ="20010-05-02";改成$staff_join ="2010-05-02"; <?$year = 5;$staff_join = '2008-05-02';$arrdate= explode('-',$staff_join);$arrdate[0] += $year;$staff_out = implode('-',$arrdate);echo $staff_out; //2013-05-02echo "<br>";echo $arrdate[0]; //2013?> <pre><?php$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> <?php $value = 5;$year = date('Y');$year += $value;echo $year.date('-m-d'); $year = 5; $staff_join = '2008-05-02';$Begin = date("Y-m-d", strtotime("+$year year $staff_join)); 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;?> <?php $year = 5; $staff_join = '2008-05-02'; $Begin = date("Y-m-d", strtotime("+$year year,$staff_join")); echo $Begin;?> 老兄去研究一下strtotime函数,手册上有详细说明。 jquery + ajax +php 实现上传图片问题~ php 备份数据库的编码问题 急!!!给些建议 求前辈们介绍个好用的PHP产品库程序 求解密此文件!!附件下载地址已附上!! 新浪授权OAUTH2.0回调页面callback为什么只能是一个页面!!! phpexcel类怎么在原来的excel表中追加数据。求解。 微信发送位置,怎么获取到周边的酒店? 要做微信服务号开发,推荐下开源框架 有没有更好的方法解决 php --with-mysql编译问题 有没有只使用cookie的论坛源码?
<?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);
?>
$staff_join ="20010-05-02";
改成$staff_join ="2010-05-02";
<?
$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
?>
<?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>
<?php
$value = 5;
$year = date('Y');
$year += $value;
echo $year.date('-m-d');
$staff_join = '2008-05-02';$Begin = date("Y-m-d", strtotime("+$year year $staff_join));
<?php
$year = 5;
$staff_join = '2008-05-02'; $Begin = date("Y-m-d", strtotime("+$year year $staff_join"));
echo $Begin;
?>
$year = 5;
$staff_join = '2008-05-02'; $Begin = date("Y-m-d", strtotime("+$year year,$staff_join"));
echo $Begin;
?>