PHP - 使用日期找出夏令时 [英] PHP - using date to find out daylight savings time

问题描述

我想给date（）一个日期，如果该日期是DST，则返回。在Windows 7框上通过xampp运行PHP 5.4.7。运行PHP 5.2.8的Linux操作系统返回0，无论我提供什么日期。

I want to give date() a date and have it return if that date is DST or not. Running PHP 5.4.7 via xampp on a windows 7 box. A linux box running PHP 5.2.8 returns 0 no matter what date I give it.

我的代码有什么问题？

echo date('I', strtotime('30-Mar-2013'));
# should return a 1 but returns 0

echo date('I', strtotime('30-Mar-2013 America/Los_Angeles'))."<br>"; # returns 0
echo date('I', strtotime('31-Mar-2013 America/Los_Angeles'))."<br>"; # returns 1

DST之间的切换应在2013年3月9日至2013年3月10日。

The switch between DST should be between 9-Mar-2013 - 10-Mar-2013.

此时，问题是为什么我的PHP代码不返回1。

At this point, the question is, why doesn't my PHP code return 1.

推荐答案

您不能使用 strtotime ，因为它创建了一个UTC时间戳记，可以删除时区信息。相反，只需使用 DateTime

You can't use strtotime because it creates a UTC timestamp, which removes the timezone information. Instead, just use DateTime

$date = new DateTime('30-Mar-2013 America/Los_Angeles');
echo $date->format('I');
# 1

$date = new DateTime('31-Mar-2013 America/Los_Angeles');
echo $date->format('I');
# 1

$date = new DateTime('09-Mar-2013 America/Los_Angeles');
echo $date->format('I');
# 0

$date = new DateTime('10-Mar-2013 America/Los_Angeles');
echo $date->format('I');
# 0

请注意，最后一个DST仍然为0？这是因为转换发生在凌晨2:00：

Notice that the DST is still 0 on that last one? That's because the transition happens at 2:00 AM:

$date = new DateTime('10-Mar-2013 01:00 America/Los_Angeles');
echo $date->format('I');
# 0

$date = new DateTime('10-Mar-2013 02:00 America/Los_Angeles');
echo $date->format('I');
# 1

这是一个弹跳式转换，时钟从2:00至3:00。

This is a "spring-forward" transition, where the clock jumps from 2:00 to 3:00.

在回退过渡期间小心模糊，时钟从2:00返回到1:00

Beware of ambiguity during fall-back transitions, where the clock jumps from 2:00 back to 1:00

$date = new DateTime('3-Nov-2013 01:00 America/Los_Angeles');
echo $date->format('I');
# 1

有两个1:00 AM，这是第一个。这是不明确的，因为我们可能意味着第二个，而不是代表。

There are two 1:00 AMs, and this is taking the first one. It is ambiguous, because we might have meant the second one, and that is not represented.

人们会认为PHP将允许以下任一项：

One would think that PHP would allow either of the following:

new DateTime('3-Nov-2013 01:00 -0700 America/Los_Angeles')  #PDT
new DateTime('3-Nov-2013 01:00 -0800 America/Los_Angeles')  #PST

但这些不我测试时似乎工作。我也尝试过ISO格式的日期。任何人都知道如何正确区分PHP中的模糊值？如果是，请在评论中编辑或更新。谢谢。

But these don't seem to work when I tested. I also tried ISO format dates. Anyone know how to properly distinguish ambiguous values in PHP? If so, please edit or update in comments. Thanks.

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