本文介绍了与leap年的日期差的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我有5个不同的时间表,为期5周:
I have 5 different schedules for 5 weeks:
- 第一周=周一至周五(上午8点至下午5点)&&星期六和星期日的休息日
- 第二周=星期一至星期五(上午10点至下午6点)及周六和周日的休息日
- 第三周=周一至周五(上午11点至下午7点)&&星期六和星期日的休息日
- 第四周=星期一休息日&&周二至周六(上午10:30至下午6:30)及星期日休息日
- 第五周=星期一休息日&&周二至周六(上午8:30至下午5:30)及周日休息日
- first week = Monday to Friday (8am to 5pm) && Rest days on Saturday and Sunday
- second week = Monday to Friday (10am to 6pm) && Rest days on Saturday and Sunday
- third week = Monday to Friday (11am to 7pm) && Rest days on Saturday and Sunday
- fourth week = Monday Rest Day && Tuesday to Saturday (10:30 am to 6:30pm) && Sunday Rest Day
- fifth week = Monday Rest Day && Tuesday to Saturday (8:30 am to 5:30pm) && Sunday Rest Day
根据我的计算数组[0],[0],即第一周的星期一设置为4月25日, 2011。
Base on my calculation array [0],[0] which is Monday of first week is set to April 25, 2011.
我有这段代码可以计算输入日期和开始日期(即2011年4月25日)之间的差。
I have this code to compute the difference between input date and start date, which is April 25, 2011.
$tdays = floor((strtotime($date2) - strtotime($date1))/86400);
我现在可以计算从2011年4月到2012年2月的工作时间表。
但是,如果我输入的日期是2012年2月以后的日期,则由于leap年而导致输出错误。
I could now compute my work schedule starting April of 2011 up until February of 2012.However if I enter a date beyond February 2012, the output is wrong due to leap year. Is there a technique for this?
推荐答案
如果您能够使用php 5.3,则应该使用
If you are able to make use of php 5.3 you should use date_diff()
或尝试像这样的东西:
<?php
function dateDifference($startDate, $endDate)
{
$startDate = strtotime($startDate);
$endDate = strtotime($endDate);
if ($startDate === false || $startDate < 0 || $endDate === false || $endDate < 0 || $startDate > $endDate)
return false;
$years = date('Y', $endDate) - date('Y', $startDate);
$endMonth = date('m', $endDate);
$startMonth = date('m', $startDate);
// Calculate months
$months = $endMonth - $startMonth;
if ($months <= 0) {
$months += 12;
$years--;
}
if ($years < 0)
return false;
// Calculate the days
$offsets = array();
if ($years > 0)
$offsets[] = $years . (($years == 1) ? ' year' : ' years');
if ($months > 0)
$offsets[] = $months . (($months == 1) ? ' month' : ' months');
$offsets = count($offsets) > 0 ? '+' . implode(' ', $offsets) : 'now';
$days = $endDate - strtotime($offsets, $startDate);
$days = date('z', $days);
return array($years, $months, $days);
}
?>
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