How to extract only 'Day' value from full 'Date' string?

问题

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?

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.

回答1:

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

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

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");

回答2:

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

Demo, docs on explode().

Or even

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

Demo, docs on substr() and strrpos().

Disclaimer: Both are oldschool!

回答3:

Try using date_parse_from_format, e.g.:

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

Testing in shell:

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

回答4:

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5:

<?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6:

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

回答7:

$dt = "2008/02/23";

echo 'The day is '. date("d", strtotime($dt));

来源：https://stackoverflow.com/questions/4275599/how-to-extract-only-day-value-from-full-date-string

标签

php

date