计算PHP中两个日期之间的工作时间

问题描述

我有一个函数可以返回两个日期之间的差异，但是我需要计算出工作时间的差异，假设是周一到周五(上午 9 点到下午 5:30):

I have a function to return the difference between 2 dates, however I need to work out the difference in working hours, assuming Monday to Friday (9am to 5:30pm):

//DATE DIFF FUNCTION
// Set timezone
date_default_timezone_set("GMT");

// Time format is UNIX timestamp or
// PHP strtotime compatible strings
function dateDiff($time1, $time2, $precision = 6) {
    // If not numeric then convert texts to unix timestamps
    if (!is_int($time1)) {
        $time1 = strtotime($time1);
    }
    if (!is_int($time2)) {
        $time2 = strtotime($time2);
    }

    // If time1 is bigger than time2
    // Then swap time1 and time2
    if ($time1 > $time2) {
        $ttime = $time1;
        $time1 = $time2;
        $time2 = $ttime;
    }

    // Set up intervals and diffs arrays
    $intervals = array('year','month','day','hour','minute','second');
    $diffs = array();

    // Loop thru all intervals
    foreach ($intervals as $interval) {
        // Set default diff to 0
        $diffs[$interval] = 0;
        // Create temp time from time1 and interval
        $ttime = strtotime("+1 " . $interval, $time1);
        // Loop until temp time is smaller than time2
        while ($time2 >= $ttime) {
            $time1 = $ttime;
            $diffs[$interval]++;
            // Create new temp time from time1 and interval
            $ttime = strtotime("+1 " . $interval, $time1);
        }
    }

    $count = 0;
    $times = array();
    // Loop thru all diffs
    foreach ($diffs as $interval => $value) {
        // Break if we have needed precission
        if ($count >= $precision) {
            break;
        }
        // Add value and interval 
        // if value is bigger than 0
        if ($value > 0) {
            // Add s if value is not 1
            if ($value != 1) {
                $interval .= "s";
            }
            // Add value and interval to times array
            $times[] = $value . " " . $interval;
            $count++;
        }
    }

    // Return string with times
    return implode(", ", $times);
}

日期 1 = 2012-03-24 03:58:58
日期 2 = 2012-03-22 11:29:16

是否有一种简单的方法可以做到这一点，即 - 计算一周工作时间的百分比并使用上述函数除以差异 - 我已经尝试过这个想法并得到了一些非常奇怪的数字......

Is there a simple way of doing this, i.e - calculating the percentage of working hours in a week and dividing the difference using the above function - I have played around with this idea and got some very strange figures...

或者有更好的方法......?

Or is there better way....?

推荐答案

这个例子使用 PHP 的内置 DateTime 类来进行日期计算.我的处理方法是首先计算两个日期之间的完整工作日数，然后将其乘以 8(见注释).然后它获取部分日期的工作时间并将它们添加到总工作时间中.把它变成一个函数会很简单.

This example uses PHP's built in DateTime classes to do the date math. How I approached this was to start by counting the number of full working days between the two dates and then multiply that by 8 (see notes). Then it gets the hours worked on the partial days and adds them to the total hours worked. Turning this into a function would be fairly straightforward to do.

注意事项:

• 不考虑时间戳.但您已经知道如何做到这一点.
• 不处理假期.(可以通过使用一系列假期并将其添加到过滤掉星期六和星期日的位置来轻松添加).
• 需要 PHP 5.3.6+
• 假设每天工作 8 小时.如果员工不吃午饭，请将 $hours = $days * 8; 更改为 $hours = $days * 8.5;

.

<?php
// Initial datetimes
$date1 = new DateTime('2012-03-22 11:29:16');
$date2 = new DateTime('2012-03-24 03:58:58');

// Set first datetime to midnight of next day
$start = clone $date1;
$start->modify('+1 day');
$start->modify('midnight');

// Set second datetime to midnight of that day
$end = clone $date2;
$end->modify('midnight');

// Count the number of full days between both dates
$days = 0;

// Loop through each day between two dates
$interval = new DateInterval('P1D');
$period = new DatePeriod($start, $interval, $end);
foreach ($period as $dt) {
    // If it is a weekend don't count it
    if (!in_array($dt->format('l'), array('Saturday', 'Sunday'))) {
        $days++;
    }
}

// Assume 8 hour workdays
$hours = $days * 8;

// Get the number of hours worked on the first day
$date1->modify('5:30 PM');
$diff = $date1->diff($start);
$hours += $diff->h;

// Get the number of hours worked the second day
$date1->modify('8 AM');
$diff = $date2->diff($end);
$hours += $diff->h;

echo $hours;

查看实际操作

参考

• 日期时间类
• DatePeriod 类
• DateInterval 类

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