PHP Regex to remove everything but the date from filename

Question

I have a filename with a date in it, the date is always at the end of the filename. And there is no extension (because of the basename function i use).

What i have:

$file = '../file_2012-01-02.txt';
$file = basename('$file', '.txt');
$date = preg_replace('PATTERN', '', $file);

Im really not good at regex, so could someone help me with getting the date out of the filename.

Thanks

Answer 1

I suggest to use preg_match instead of preg_replace:

$file = '../file_2012-01-02';
preg_match("/.*([0-9]{4}-[0-9]{2}-[0-9]{2}).*/", $file, $matches);
echo $matches[1]; // contains '2012-01-02'

Answer 2

I suggest you to try:

$exploded = explode("_", $filename);
echo $exploded[1] . '<br />'; //prints out 2012-01-02.txt
$exploded_again = explode(".", $exploded[1]);
echo $exploded_again[0]; //prints out 2012-01-02

Shorten it:

$exploded = explode( "_" , str_replace( ".txt", "", $filename ) );
echo $exploded[1];

Answer 3

If there is always an underscore before the date:

ltrim(strrchr($file, '_'), '_');
      ^^^^^^^ get the last underscore of the string and the rest of the string after it
^^^^^ remove the underscore

Answer 4

这样，当您确实需要使用regexp时：

current(explode('.', end(explode('_', $filename))));

Answer 5

This should help i think:

<?php

$file = '../file_2012-01-02.txt';
$file = basename("$file", '.txt');
$date = preg_replace('/(\d{4})-(\d{2})-(\d{2})$/', '', $file);

echo $date; // will output: file_

?>