PHP，使用 strtotime 从日期时间变量中减去分钟？
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PHP, use strtotime to subtract minutes from a date-time variable?
提问by Klausos Klausos
I need to subtract 45 minutes from the date-time variable in PHP.
我需要从 PHP 中的日期时间变量中减去 45 分钟。
The code:
编码：
$thestime = '2012-07-27 20:40';
$datetime_from = date("Y-m-d h:i",strtotime("-45 minutes",strtotime($thestime)));
echo $datetime_from;
returns the result 2012-07-27 07:55.
返回结果2012-07-27 07:55。
It should be 2012-07-27 19:55, though. How do I fix this?
2012-07-27 19:55不过应该是的。我该如何解决？
回答by Bogdan
You should do:
你应该做：
$datetime_from = date("Y-m-d H:i", strtotime("-45 minutes", strtotime($thestime)));
Having Hinstead of hmeans a 24-hour format is used, representing the hour with leading zeros: 00through 23.
有H而不是h意味着使用 24 小时格式，用前导零表示小时：00到23。
You can read more on this in the PHP date function documentation.
您可以在PHP 日期函数文档中阅读更多相关内容。
There are also object oriented ways of doing this which are more fluent, like DateTime::sub:
还有一些面向对象的方法可以更流畅地执行此操作，例如DateTime::sub：
$datetime_from = (new DateTime($thestime))->sub(DateInterval::createFromDateString('45 minutes'))->format('Y-m-d H:i')
Or the even more expressive way offered by the Carbonlibrary which extends PHP's built in DateTimeclass:
或者Carbon扩展 PHP 内置DateTime类的库提供的更具表现力的方式：
$datetime_from = (new Carbon($thestime))->subMinutes(45)->format('Y-m-d H:i');
