问题描述

当时间大于 24 小时时，计算时间差的最佳方法是什么.

示例

$time1 = '76:00:00';$time2 = '30:00:00';//结果应该是 46:00:00echo date('H:i:s', strtotime($time1) - strtotime($time2));

但这不能用这个来完成，因为它大于 24 小时.

我还在数据库中存储了这样的时间:33:30:00我如何在 php 中将其格式化为:33:30

使用DateTime和DateInterval进行计算:

$date1 = new DateTime('now', new DateTimeZone('UTC'));$date2 = new DateTime('now', new DateTimeZone('UTC'));$time1 = new DateInterval('PT76H');$time2 = new DateInterval('PT30H');$date1->add($time1);$date2->add($time2);$diff = $date1->diff($date2);回声 ($diff->days * 24 + $diff->h) .$diff->format(':%I:%S');

说明:无法直接在 DateInterval 上执行计算，因此您必须创建日期作为计算的基础.然后将两个不同的间隔添加到当前日期，并计算它们之间的差异.diff() 返回包含总天数的 DateInterval，您必须乘以 24 才能获得小时数，以及不等于完整天数的小时数.p>

时区应指定为 UTC 以避免夏令时问题.

What is the best way to calculate the difference in time, when time is greater than 24 hours.

Example

$time1 = '76:00:00';
$time2 = '30:00:00';

// result should be 46:00:00
echo date('H:i:s', strtotime($time1) - strtotime($time2));

But this could not be done with this because its greater then 24 hours.

Also in a database i've stored a time like this: 33:30:00 How in php could i format it to: 33:30

Use DateTime and DateInterval to perform calculations:

$date1 = new DateTime('now', new DateTimeZone('UTC'));
$date2 = new DateTime('now', new DateTimeZone('UTC'));
$time1 = new DateInterval('PT76H');
$time2 = new DateInterval('PT30H');

$date1->add($time1);
$date2->add($time2);

$diff = $date1->diff($date2);
echo ($diff->days * 24 + $diff->h) . $diff->format(':%I:%S');

Explanation: It's not possible to perform calculations directly on DateIntervals, so you have to create dates as a basis for calculations. Then add two different intervals to current dates, and calculate a difference between them. diff() returns DateInterval that contains total number of days, that you have to multiply by 24 to get hours, and hours that don't make full days.

EDIT: The timezone should be specified as UTC to avoid daylight saving time issues.

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