php 如何从完整的“日期”字符串中仅提取“天”值？
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How to extract only 'Day' value from full 'Date' string?
提问by Hyman Roscoe
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?
我有一个格式为'Y-m-d'. 我想遍历数组，只提取'd'每个成员的日期 ( )，但无法弄清楚如何拆分它。我是否以某种方式使用子字符串？
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.
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或两位数。
回答by Pekka
If you have PHP < 5.3, use strtotime():
如果您的 PHP < 5.3，请使用strtotime()：
$string = "2010-11-24";
$timestamp = strtotime($string);
echo date("d", $timestamp);
If you have PHP >= 5.3, use a DateTimebased solution using createFromFormat- DateTimeis the future and can deal with dates beyond 1900 and 2038:
如果您的 PHP >= 5.3，请使用DateTime基于createFromFormat的解决方案-DateTime是未来，可以处理 1900 年和 2038 年以后的日期：
$string = "2010-11-24";
$date = DateTime::createFromFormat("Y-m-d", $string);
echo $date->format("d");
回答by jensgram
回答by kenorb
Try using date_parse_from_format, e.g.:
尝试使用date_parse_from_format，例如：
date_parse_from_format('Y-m-d', '2010-11-24')['day'];
Testing in shell:
在外壳中测试：
$ php -r "echo date_parse_from_format('Y-m-d', '2010-11-24')['day'];"
24
回答by Marek Augustyn
php> $d = '2010-11-24';
'2010-11-24'
php> $fields = explode('-', $d);
array (
0 => '2010',
1 => '11',
2 => '24',
)
php> $fields[2];
'24'
回答by martynthewolf
<?php
$dates = array('2010-01-01', '2010-01-02', '2010-01-03', '2010-01-23');
foreach ($dates as $date) {
$day[] = substr($date, 8, 2);
}
var_dump($day);
?>
回答by Frank Mu?oz
$string = "2010-11-24";
$date = new DateTime($string);
echo date_format($date, 'd');
回答by njoshsn
$dt = "2008/02/23";
$dt = "2008/02/23";
echo 'The day is '. date("d", strtotime($dt));
echo '今天是'。日期("d", strtotime($dt));
