Take if there is an integer part as the tail of a string variable in PHP

Question

How to extract the integer part(the last integer) of a string variable to another integer variable in PHP.

Here is the current status of my code:

<?php
 $path = current_path(); // eg. loading 'node/43562' to $path
 $pos = strpos($path, 'node/');
 if($pos>-1)die(print $pos); // sometimes $pos can be 0, may be 5 or even 26..like so
 ...
?>

All I need is to get the integer part comes after 'node/' out from $path only if the integer is the last thing in the $part.

Do nothing if $path is:

'greenpage/12432/node', or
'56372/page/2321', or
'node/56372/page'.

I don't want to go through lengthy codes like:

<?php
 $arr = explode(":",$path);
 $a= $arr[1];
 $b= $arr[3];
 $c= $arr[5];
 ...
 ...
?>

I have to use this 43562 in many places. Hope it can be implemented with any simple preg_ methods, or a complex regex.

Waiting for the minimal LOC solution.

writemycodeforme.com

Lightness Races in Orbit
– Lightness Races in Orbit

2015-01-02 05:08:34 +00:00
Commented Jan 2, 2015 at 5:08 — Lightness Races in Orbit
– Lightness Races in Orbit, Commented Jan 2, 2015 at 5:08

Sverri M. Olsen · Accepted Answer · 2015-01-02 05:03:52Z

1

You can get to the number in a myriad of ways.

Here is a simple regex:

preg_match('/\d+$/', $path, $match);
var_dump($match);

If it is important to check that the prefix is also there then:

preg_match('/(?<=node\/)\d+$/', $path, $match);
var_dump($match);

You can also use strrpos() and substr():

$slash = strrpos($path, '/');
if ($slash !== false) {
    echo substr($path, $slash + 1);
}

It is also possible to just chop up the string. It is a little quick'n'dirty, but it works:

$parts = explode('/', $path);
echo array_pop($parts);

edited Jan 2, 2015 at 5:03

answered Jan 2, 2015 at 4:58

Sverri M. Olsen

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Comments

Kevin · Accepted Answer · 2015-01-02 04:58:37Z

1

$string = "foo/bar/node/1234";

if(preg_match('/node\/([0-9]+)$/', $string, $match)) {

        echo $match[1];
}

answered Jan 2, 2015 at 4:58

Kevin

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1 Comment

Barmar Over a year ago

You mean $match[1], don't you?

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