如何从完整的“日期”字符串中只提取“day”值?

如果您有php<5.3,请使用 strtotime() :

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

如果php>=5.3,请使用 DateTime 基于解决方案使用 createFromFormat - 日期时间 是未来,可以处理1900年和2038年之后的日期:

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

$date = '2010-11-24';
list($y, $m, $d) = explode('-', $date);

Demo ,文档打开 explode() .

甚至

$date = '2010-11-24';
$d = substr($date, strrpos($date, '-') + 1);

Demo ,文档打开 substr() 和 strrpos() .

免责声明:两个都是老派!

php> $d = '2010-11-24';
'2010-11-24'
php> $fields = explode('-', $d);
array (
  0 => '2010',
  1 => '11',
  2 => '24',
)
php> $fields[2];
'24'

<?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);

?>

$string = "2010-11-24";
$date = new DateTime($string);
echo date_format($date, 'd');

尝试使用 date_parse_from_format 例如:

date_parse_from_format('Y-m-d', '2010-11-24')['day'];

壳牌公司测试:

$ php -r "echo date_parse_from_format('Y-m-d', '2010-11-24')['day'];"
24

$dt=“2008/02/23”;

回音“今天是”。日期(“d”,strtotime($dt));