php date_diff以小时为单位 [英] php date_diff in hours

问题描述

如何使代码下面的转换天数小时？

How is it possible to make the code below convert days in hours?

$timestart = date_create('02/11/2011' . $row->timestart); //$row->timestart returns time in 00:00:00 format
$timestop = date_create('02/11/2011' . $row->timestop); //$row->timestop returns time in 00:00:00 format

date_add($timestop, date_interval_create_from_date_string('2 days')); //add 2 days

$date_diff = date_diff($timestart, $timestop);

echo "Timespan: ";
echo $date_diff->format('%h hours');
echo "<br />";

如何获取小时：分钟：秒已经过去了我正在努力留下 date_diff 函数。

How can I get the hours:minutes:seconds elapsed? I'm trying to stay with the date_diff function.

推荐答案

date_diff（）的结果是 DateInterval 类的对象。这样的对象具有非常有用的属性 - $ days ：它是开始和结束日期之间的总天数。此外，它存储（作为其公共属性）小时，分钟和秒的差异。

The result of date_diff() is an object of DateInterval class. Such object has a very useful property - $days: it's total number of days between the starting and the ending dates. Besides, it stores (as its public properties) the difference in hours, minutes and seconds.

所以，我想，你需要的只是提取这些属性的值$ date_diff变量，然后将 24 * $ days 添加到小时数。 ）所有这些都可以包装成一个简单的函数：

So, I suppose, what you need is just extract values of these properties from $date_diff variable, then add 24*$days to the hours number. ) All this can be wrapped into a simple function:

function hms_date_diff(DateInterval $date_diff) {
  $total_days = $date_diff->days;
  $hours      = $date_diff->h;
  if ($total_days !== FALSE) {
    $hours += 24 * $total_days;
  }
  $minutes    = $date_diff->i;
  $seconds    = $date_diff->s;
  return sprintf('%02d:%02d:%02d', $hours, $minutes, $seconds);
}

对于DateDiff :: format， doc 说...

As for DateDiff::format, the doc says...

DateInterval :: format（）方法不会重新计算在时间字符串或日期段中的
点。

The DateInterval::format() method does not recalculate carry over points in time strings nor in date segments.

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