問題描述

我有一個函數可以返回兩個日期之間的差異，但是我需要計算出工作時間的差異，假設是周一到周五(上午 9 點到下午 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

是否有一種簡單的方法可以做到這一點，即 - 計算一周工作時間的百分比并使用上述函數除以差異 - 我已經嘗試過這個想法并得到了一些非常奇怪的數字......

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....?

推薦答案

這個例子使用 PHP 的內置 DateTime 類來進行日期計算.我的處理方法是首先計算兩個日期之間的完整工作日數，然后將其乘以 8(見注釋).然后它獲取部分日期的工作時間并將它們添加到總工作時間中.把它變成一個函數會很簡單.

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.

注意事項:

• 不考慮時間戳.但您已經知道如何做到這一點.
• 不處理假期.(可以通過使用一系列假期并將其添加到過濾掉星期六和星期日的位置來輕松添加).
• 需要 PHP 5.3.6+
• 假設每天工作 8 小時.如果員工不吃午飯，請將 $hours = $days * 8; 更改為 $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;

查看實際操作

參考

• 日期時間類
• DatePeriod 類
• DateInterval 類

這篇關于計算PHP中兩個日期之間的工作時間的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！