如何减去两个日期，忽略PHP中的夏令时？ [英] How to subtract two dates, ignoring daylight savings time in PHP?

问题描述

我试图计算两天之间的天数，但是我遇到了夏令时间的问题。这是我的代码：

I'm trying to calculate the number of days between two days, but I'm running into issues with Daylight Savings Time. Here's my code:

function date_diff($old_date, $new_date) {
    $offset = strtotime($new_date) - strtotime($old_date);
    return $offset/60/60/24;
}

只要日期都在相同的DST期间， / p>

Works fine as long as the days are both within the same DST period:

echo date_diff('3/15/09', '3/18/09'); // 3

但是如果他们分开，则不是：

But not if they're further apart:

echo date_diff('11/15/08', '3/18/09'); // 122.95833333333

我想要几天，而不在乎DST。我想我可以舍弃结果，但是感觉到k。。有更好的（容易）的方式吗？我不想写一整天的解析和计数避免闰年的东西，如果我可以避免。

I want an even number of days, and don't care about DST. I suppose I could round the result, but that feels kludgy. Is there a better (easy) way? I don't want to have to write a whole day-parsing-and-counting-avoiding-leap-years thing if I can avoid it.

（注意：这个必须在php 5.1.6下运行，所以5.3中的某些日期功能可能无法使用。）

(Note: this has to run under php 5.1.6, so some of the date features in 5.3 may not be available.)

更多信息：我要采取偏移并将其添加到数据库中的其他数据时间，我只想要更改日期部分，而不是时间部分。反正四舍五入不行，无论如何，因为当我加入时，在另一个方向上下了一个小时。也许对整个问题有一个更好的方法....

A bit more info: I'm going to take the offset and add it to other datetimes that are in a db, and I want only the day part to change, not the time part. Turns out rounding won't work, anyway, because when I do the adding it gets off by one hour in the other direction. Maybe there's a better approach to the whole problem....

推荐答案

你可以使用 http://ca3.php.net/date_default_timezone_set 将时区设置为GMT，所以不会有偏移。

you could use http://ca3.php.net/date_default_timezone_set to set the timezone to GMT so there will be no offset.

另外，您可以使用日期（'I'，$ timetamp）手动添加一个偏移量

if ( date("I") == 1 ) {      // "WE ARE MDT";  
        $timeZone = "MDT";  
} else {  
        $timeZone = "MST";  
}  

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