How get last segment of url using PCRE?

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

如果可以，请对PCRE拒绝：-)。

echo basename('http://example.com/sales/view/id/705');

Answer 7

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";

Answer 8

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