PHP计算时间差大于24小时 | PHP计算时间差大于24小时

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问题描述

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

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

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

但这不能完成，因为它超过24小时。

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

另外在数据库中我已经存储了一个这样的时间：33:30:00
如何将PHP格式化为：33:30

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

推荐答案

使用 \DateTime 和 \DateInterval 执行计算：

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

说明：
无法直接在 DateInterval s，所以你必须创建日期作为计算的基础。然后为当前日期添加两个不同的间隔，并计算它们之间的差异。 diff（）返回 \DateInterval ，其中包含总天数，您必须乘以24才能获得小时数，并且没有整整一天的时间。

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.

编辑：时区应指定为UTC，以避免夏令时问题。

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

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