使用PHP计算营业日：这是准确的吗？

Question

概述：

我正在尝试计算我正在构建的自定义项目管理系统的两个日期之间的工作日。这是我到目前为止所做的。它似乎工作正常但我不确定它是多么准确以及我是否应该使用它。任何反馈都将非常感谢！

CODE

<?php

    # date variables:
    date_default_timezone_set('Asia/Kuwait');
    $date['start'] = date('Y-m-d H:i:s');
    $date['end'] = '2013-12-01 08:00:00';
    $date['off'] = array('Friday','Saturday'); # usual days off in Kuwait)

    # calculate the difference:
    $date_s = new DateTime($date['start']);
    $date_e = new DateTime($date['end']);
    $interval = $date_s->diff($date_e);

    # obtain relevant values:
    $remaining_days = $interval->format('%r%a');
    $remaining_weeks = floor($remaining_days/7);
    $weekend_days = ($remaining_weeks*count($date['off']));

    # additional holidays (just an example):
    $extra_holidays_array = array
    (
        'holiday 1' => '2013-06-24 00:00:00',
        'holiday 2' => '2013-06-25 00:00:00',
        'holiday 3' => '2013-07-01 00:00:00',
        'holiday 4' => '2013-08-24 00:00:00'
    );

    # check if a real holiday:
    $extra_holidays = 0;
    foreach( $extra_holidays_array as $check_date )
    {
        $day_of_holiday = date('l', strtotime($check_date));
        if( ! in_array($day_of_holiday,$date['off']) ){ $extra_holidays++; }
    }

    # total holidays:
    $total_holidays = ($weekend_days+$extra_holidays);

    # business days:
    $business_days_nh = ($remaining_days-$weekend_days); # NO extra holidays
    $business_days_wh = ($remaining_days-$weekend_days-$extra_holidays); # WITH extra holidays

?>

<ul>
<li>Current Date: <?php echo $date['start']; ?></li>
<li>Deadline Date: <?php echo $date['end']; ?></li>
</ul>
<ul>
<li>Remaining Days: <?php echo $remaining_days; ?></li>
<li>Remaining Weeks: <?php echo $remaining_weeks; ?></li>
</ul>
<ul>
<li>Usual Holidays: <?php echo $weekend_days; ?></li>
<li>Extra Holidays: <?php echo $extra_holidays; ?></li>
<li>Total Holidays: <?php echo $total_holidays; ?></li>
</ul>
<ul>
<li>Business Days (Before Holidays): <?php echo $business_days_nh; ?></li>
<li>Business Days (After Holidays): <?php echo $business_days_wh; ?></li>
</ul>

Answer 1

为了这个目的，我在之前的项目中编写了一个函数：

以下是用作参数的内容：

• $ start - 启动Unix时间戳（您可以使用strtotime（'2013年1月1日'）来获取时间戳
• $ end - 结束unix时间戳
• 可选$ holidays - 计算为假期的日期数组（IE银行假期等），它使用strtotime()转换日期
• 可选$ returnAsArray - 返回工作日时间戳列表，或只是工作天数

重要 重要的是要知道，给出的时间戳应该是当天开始的时间戳，例如：

要将2013年1月1日用作时间戳，请使用以下值：mktime(0, 0, 0, 1, 1, 2013);

function networkDays($start, $end, array $holidays = array(), $returnAsArray = false)
{
    if(!is_int($start)){ 
        trigger_error('Parameter 1 expected to be integer timestamp. ' . ucfirst(gettype($start)) . ' given.', E_USER_WARNING);
        return false;
    }

    if(!is_int($end)){ 
        trigger_error('Parameter 2 expected to be integer timestamp. ' . ucfirst(gettype($start)) . ' given.', E_USER_WARNING);
        return false;
    }

    if(!is_array($holidays)){ 
        $holidays = array();
        trigger_error('Parameter 3 expected to be Array. ' . ucfirst(gettype($holidays)) . ' given.', E_USER_NOTICE);
    }

    if(!is_bool($returnAsArray)){ 
        trigger_error('Parameter 4 expected to be Boolean. ' . ucfirt(gettype($returnAsArray)) . ' given.', E_USER_WARNING);
        return false;
    }


    if($start>=$end){ 
        $nEnd = $start;
        $nStart = $end;

        $start = $nStart;
        $end = $nEnd;
    }


    foreach($holidays as $key => $holiday)
    {
        $holidays[$key] = strtotime($holiday);
    }


    $numberOfDays = ceil(((($end-$start)/60)/60)/24);

    $networkDay = 0;
    $networkDayArray = array();

    for($d = 0; $d < $numberOfDays; $d++)
    {
        $dayTimestamp = $start+(86400*$d);

        if(date('N',$dayTimestamp)<6 && !in_array($dayTimestamp,$holidays))
        {
            $networkDay += 1;
            $networkDayArray[] = $dayTimestamp;
        }
    }


    if($returnAsArray)
    {
        return $networkDayArray;
    } else {
        return $networkDay;
    }
}