后端开发|php教程
nbsp,time,date,needed,end
后端开发-php教程
请问这种计算工作时间结束的具体时间戳 应该怎么写
起始时间是随机的日期时间戳是X
总共需要工作6天才能完成
但是上班时间只有5天/周
我想得到下周几才能完成，包括具体的时间戳，星期几
可生成积分卡php源码,ubuntu设置60帧,tomcat怎么导入互联网,java爬虫解析dom,php8新特性之match,seo媒体发稿lzw
这里要如何判断周六日，把他们再每一次都隔过去呢
老罗android开发 源码,vscode提交代码对比工具,ubuntu 开发安卓,tomcat没有执行权限,sqlite金融数据,网页设计导航栏的字体,阿里云服务器ecs win,web聊天室插件,商城前端ui框架,爬虫是超,夏日php,刘毅seo,springboot埋点监控,宁志地方政府网站管理系统,简单静态网页模板下载,帝国cms 自媒体模板,建站后台新建文章不显示,dedecms 页面跳转,网站的管理系统,小程序视频软件有哪些lzw
麻烦写个完整代码，感谢
——解决思路———————-
网站源码 下载,ubuntu取词发音翻译,水生透明小爬虫,php $limit,seo网站审查lzw
<?php
$start_time =rand(time(), time() + 50 * 24 * 60 * 60);
$needed_time = 23; //day
$work_day = 5;
$needed_week = floor($needed_time/$work_day);
$end_time = $start_time + $needed_week * 7 * 24 * 60 * 60 + $needed_time%$work_day * 24 * 60 * 60;
$end_date = date("Y-m-d", $end_time);
$start_date = date("Y-m-d", $start_time);
$end_week_day = date("w", $end_time);
$weeks = array("Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday");
echo "start date: $start_date, end date: $end_date, " . $weeks[$end_week_day] . " after ". $needed_week . " weeks.";
——解决思路———————-
<?php
$start_time =rand(time(), time() + 50 * 24 * 60 * 60);
$needed_time = 23; //day
$work_day = 5;
$needed_week = floor($needed_time/$work_day);
$end_time = $start_time + $needed_week * 7 * 24 * 60 * 60 + $needed_time%$work_day * 24 * 60 * 60;
$weeks = array("Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday");
$end_week_day = date("w", $end_time);
if($end_week_day == 6
------解决思路----------------------
$end_week_day == 0)
{
$needed_week ++;
$end_time += 2 * 24 * 60 * 60;
}
$end_date = date("Y-m-d", $end_time);
$start_date = date("Y-m-d", $start_time);
$end_week_day = date("w", $end_time);
echo "start date: $start_date, end date: $end_date, " . $weeks[$end_week_day] . " after ". $needed_week . " weeks.";
——解决思路———————-
与其费脑筋推演算式，还不如用计算机最擅长简单累加
$x = -07-27;
$d = 6;
$t = strtotime($x);
while($d > 1) { //当天不算就 $d > 0
$t += 86400;
if(in_array(date(w, $t), array(6,0))) continue;
$d--;
}
echo date(Y-m-d w, $t);
-08-03 1
