PHP DateTime正则表达式 [英] PHP DateTime Regex

问题描述

长话短说，我继承了一些糟糕的代码。结果，由于日期格式的原因，在比较日期时，字符串比较是错误的。我试图将日期转换为有效的DateFormat语法，以便可以进行适当的比较。

Hey, long story short I have inherited some terrible code. As a result a string comparison is buggy when comparing dates due to the format of the date. I am trying to convert the date to a valid DateFormat syntax so I can run a proper comparison.

这些是当前格式的一些示例：

These are some samples of the current format:

12/01/10 at 8:00PM

12/31/10 at 12:00PM

12/10/09 at 5:00AM

等。我想将其转换为YYYYMMDDHHMM格式，即201012012000，以进行比较。如果有人可以给我一个快速的正则表达式片段来执行此操作，那将不胜感激，因为我现在正在为正则表达式打砖墙。我可以通过多次分解字符串来做到这一点，但是我宁愿以一种更有效的方式来做到这一点。

and so forth. I'd like to convert this to a YYYYMMDDHHMM format i.e 201012012000 for comparison purposes. If anyone can give me a quick regex snippet to do this that'd be appreciated as right now i'm hitting a brick wall for a regex. I can do it by exploding the string over several times etc but I'd rather do it in a more efficient manner.

谢谢！

推荐答案

使用 DateTime 类，它内置于PHP 5.3中。

Working with dates in strange formats is very easy with the DateTime class which was built into PHP 5.3.

不需要正则表达式或任何花哨的内容：

No need for regex or anything fancy:

$date = DateTime::createFromFormat('m/d/y \a\t g:iA', '12/10/09 at 5:00AM');
print_r($date);

一旦它是一个日期对象，就可以使用任何想要的格式。

Once it is a date object you can have it in any format you want.

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