How to return only needed match with preg_match

Question

I am trying to parse a integer from a list of uris like this:

uri.com/upload/123456789_abc.ext

I am using this pattern:

preg_match( "#uri\.com\/upload\/(.*?)_#is", $uri, $match );

Which works and returns:

Array
(
    [0] => uri.com/upload/123456789_
    [1] => 123456789
)

But I was wondering if there's a way to make $match == "123456789" intead of returning an array with multiple values.

Is it possible to do it by only modifying the pattern?

Answer 1

It will always return an array, but you can change the pattern, so that it only matches what you want.

$uri = "uri.com/upload/123456789_abc.ext";
preg_match('#(?<=uri\.com/upload/)\d+#is', $uri, $match );
print_r($match);

returns

Array ( [0] => 123456789 )

so it is still an array, but it does only contain the whole match, that is your number.

(?<=uri\\.com/upload/) is a lookbehind, it does not match that part, so it is not part of the result.

\d+ is only matching digits, so the _ is not needed anymore.