php计算给定时间段内一共有多少周(可以支持跨年计算) 例如:2012-02-04到2012-05-06间隔多少周,周一为一周的起始跨年也可以支持如2011-12-04到2012-05-06间隔多少周 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 function computeWeek($date1,$date2){ $diff = strtotime($date2) - strtotime($date1); $res = ceil($diff/(24*60*60*7)); return $res;}echo computeWeek("2011-06-06","2012-06-06")."<br>"; #53echo computeWeek("2012-06-01","2012-06-07")."<br>"; #1 function computeWeek($date1,$date2){ $diff = strtotime($date2) - strtotime($date1); $week = floor($diff/(24*60*60*7)); $day = $diff%(24*60*60*7)/(24*60*60); $res = $week."周又".$day."天"; return $res;}echo computeWeek("2011-06-06","2012-06-06")."<br>"; #52周又2天echo computeWeek("2012-06-01","2012-06-09")."<br>"; #1周又1天 function computeWeek($date1,$date2){ $diff = abs(strtotime($date2) - strtotime($date1)); #取差集的绝对值 $week = floor($diff/(24*60*60*7)); #获取多少周 $day = ($diff%(24*60*60*7))/(24*60*60); #除周数以外的天数 $res = $week."周又".($day+1)."天"; return $res;}echo computeWeek("2011-06-06","2012-06-06")."<br>"; #52周又3天echo computeWeek("2012-06-09","2012-06-01")."<br>"; #1周又2天 关于die的问题 关闭浏览器,清除cookie PHP重点基础知识整理[长期更新] 求助,看不懂 Zend Studio for Eclipse--新建项目问题 $_FILES ["file1"] ["name"]返回值代表什么 本机测试可以,但服务器上就不能上传图片了。 求PHP操作XML的代码 网站高手请进!!!欢迎交流(100分等) 这段程序怎么修改啊? 连oracle长时间后报错 关于XML的字符集的问题
function computeWeek($date1,$date2){
$diff = strtotime($date2) - strtotime($date1);
$res = ceil($diff/(24*60*60*7));
return $res;
}
echo computeWeek("2011-06-06","2012-06-06")."<br>"; #53
echo computeWeek("2012-06-01","2012-06-07")."<br>"; #1
function computeWeek($date1,$date2){
$diff = strtotime($date2) - strtotime($date1);
$week = floor($diff/(24*60*60*7));
$day = $diff%(24*60*60*7)/(24*60*60);
$res = $week."周又".$day."天";
return $res;
}
echo computeWeek("2011-06-06","2012-06-06")."<br>"; #52周又2天
echo computeWeek("2012-06-01","2012-06-09")."<br>"; #1周又1天
function computeWeek($date1,$date2){
$diff = abs(strtotime($date2) - strtotime($date1)); #取差集的绝对值
$week = floor($diff/(24*60*60*7)); #获取多少周
$day = ($diff%(24*60*60*7))/(24*60*60); #除周数以外的天数
$res = $week."周又".($day+1)."天";
return $res;
}
echo computeWeek("2011-06-06","2012-06-06")."<br>"; #52周又3天
echo computeWeek("2012-06-09","2012-06-01")."<br>"; #1周又2天