如何从完整的“日期"字符串中仅提取“天"值? [英] How to extract only 'Day' value from full 'Date' string?

问题描述

我有一个格式为'Y-m-d'的Date值数组.我想遍历数组，只提取每个成员的日期('d')，但不知道要拆分它.我会以某种方式使用子字符串吗?

I have an array of Date values in the format of 'Y-m-d'. I want to loop over the array and only extract the day ('d') of each member, but can't figure out to split it. Do I use substrings in some way?

简单地说，我有'2010-11-24'，我想从中得到 '24'.问题是一天可能是个位数还是个位数.

Put simply, I have '2010-11-24', and I want to get just '24' from that. The problem being the day could be either single or double figures.

推荐答案

如果您有PHP<5.3，使用 strtotime() :

If you have PHP < 5.3, use strtotime():

$string = "2010-11-24";
$timestamp = strtotime($string);
echo date("d", $timestamp);

如果您的PHP> = 5.3，请使用基于 DateTime 的解决方案，并使用 createFromFormat - DateTime 是将来的日期，可以处理1900年和2038年以后的日期:

If you have PHP >= 5.3, use a DateTime based solution using createFromFormat - DateTime is the future and can deal with dates beyond 1900 and 2038:

 $string = "2010-11-24";
 $date = DateTime::createFromFormat("Y-m-d", $string);
 echo $date->format("d");

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