How get last segment of url using PCRE? - php

Question

I have a url, - "http://example.com/sales/view/id/705" and I need get a last segment (705).
How can I do this using PCRE?

Answer 1

This should do it in Perl:
my ($last) = $url =~ /([^\/]+)\z/;
But I would rather use the URI module:
my $last = (URI->new($url)->path_segments)[-1];

Answer 2

(In PHP) I would not use PCRE for such a trivial and un-ambiguous job. I would just do:
$parts = explode('/', rtrim($url, '/'));
$partYouWant = array_pop($parts);
EDIT
If you need to use PCRE (although I don't know why you would) this variation on eugene y's answer would do it:
$pattern = '#/([^/]+)\z#';
$url = 'http://example.com/sales/view/id/705';
preg_match($pattern, $url, $matches);
echo $matches[1];

Answer 3

In PHP you can do this in a single line code:
$url = 'http://example.com/sales/view/id/705';
substr($url, strrpos($url, '/') + 1);

Answer 4

Non PCRE alternative:
$url="http://example.com/sales/view/id/705";
$lastPart = current(array_reverse((explode('/',parse_url($url,PHP_URL_PATH)))));
Doubt if it's any faster though

Answer 5

You could use this pattern ([^\/]*)$ for everything from last / to end.
Maybe also interesting: ([^\/\?]*)(\?.*)?$ gives you everything between last / and first ?

Answer 6

Say no to PCRE if you can,:-).
echo basename('http://example.com/sales/view/id/705');

Answer 7

preg_match('#/([1-9]\d*)/?(?:$|\?)#', $url, $matches);//$matches[1] contains your id

Answer 8

Simplest:
$ok=preg_match('#\d+$#',$url,$m);
if($ok)
echo $m[0],"\n";
Brainy:
$ok=preg_match('#/(\d+)$#',$url,$m);
if($ok)
echo $m[1],"\n";
Flexible: (as it also allows words, other than digits)
$ok=preg_match('#/(\w+)$#',$url,$m);
if($ok)
echo $m[1],"\n";
More flexible: (as it now allows everything that's not a / to match)
$ok=preg_match('#/(.*?)$#',$url,$m);
if($ok)
echo $m[1],"\n";