计算PHP中的两个日期之间的工作时间 [英] calc the working hours between 2 dates in PHP

问题描述

我有一个功能来返回2个日期之间的差异，但是我需要确定工作时间的差异，假设星期一到星期五（上午9点到下午5点30分）：

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

有没有一个简单的方法，即 - 计算一周工作时间的百分比，并使用上述功能划分差异 - 我已经玩弄了这个想法，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...

或有更好的方法....？

Or is there better way....?

推荐答案

此示例使用PHP内置的DateTime类来执行日期数学。我接近这个是从两个日期之间的整个工作日数开始，然后乘以8（见说明）。然后，它可以在部分日子工作，并将其添加到工作时间的总时间。将其变成一个功能将是相当直接的。

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：

• 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轻松添加，并将其添加到星期六和星期日过滤的地方）。


• 需要PHP 5.3.6 +

假设工作时间为8小时。如果员工不吃午餐，则$ $ hours = $ days * 8; 至 $ hours = $ days * 8.5;

• Does not take timestamps into account. But you already know how to do that.
• Does not handle holidays. (That can be easily added by using an array of holidays and adding it to where you filter out Saturdays and Sundays).
• Requires PHP 5.3.6+
• Assumes an 8 hour workday. If employees do not take lunch change $hours = $days * 8; to $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;

看到它

See it in action

参考

• DateTime Class


• DatePeriod类


• DateInterval Class

• DateTime Class
• DatePeriod Class
• DateInterval Class

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