考虑 PHP 中的日期计算持续时间

Question

考虑到 PHP 中的日期戳，我如何计算持续时间？我在日期之间使用的日期格式是“Y-m-d H:i:s”，

我的工作代码只能计算时间之间的持续时间而不考虑日期。

下面是我的代码：

$assigned_time = "2012-05-21 22:02:00";
$completed_time= "2012-05-22 05:02:00";   

function hmsDiff ($assigned_time, $completed_time) {
    $assigned_seconds = hmsToSeconds($assigned_time);
    $completed_seconds = hmsToSeconds($completed_time);

    $remaining_seconds = $assigned_seconds - $completed_seconds;

    return secondsToHMS($remaining_seconds);
}
function hmsToSeconds ($hms) {
    $total_seconds = 0;
    list($hours, $minutes, $seconds) = explode(":", $hms);
    $total_seconds += $hours * 60 * 60;
    $total_seconds += $minutes * 60;
    $total_seconds += $seconds;

    return $total_seconds;
}

 function secondsToHMS ($seconds) {
    $minutes = (int)($seconds / 60); 
    $seconds = $seconds % 60;
    $hours = (int)($minutes / 60); 
    $minutes = $minutes % 60;

    return  sprintf("%02d", abs($hours)) . ":" . 
        sprintf("%02d", abs($minutes)) . ":" . 
        sprintf("%02d", abs($seconds));

 }

Answer 1

不完全知道你想要什么，比如：

// prevents php error
date_default_timezone_set ( 'US/Eastern' );
// convert to time in seconds
$assigned_seconds = strtotime ( $assigned_time );
$completed_seconds = strtotime ( $completed_time );

$duration = $completed_seconds - $assigned_seconds;

// j gives days
$time = date ( 'j g:i:s', $duration );

Answer 2

DateTime 有一个“diff”方法，它返回一个 Interval 对象。区间对象有一个方法"format" which allows you to customize the output。

#!/usr/bin/env php
<?php

$assigned_time = "2012-05-21 22:02:00";
$completed_time= "2012-05-22 05:02:00";   

$d1 = new DateTime($assigned_time);
$d2 = new DateTime($completed_time);
$interval = $d2->diff($d1);

echo $interval->format('%d days, %H hours, %I minutes, %S seconds');

注意：如果您没有使用 5.3.0+，这里有一个很好的答案：https://stackoverflow.com/a/676828/128346。