与给定模式匹配的正则表达式，以可选数字结尾

Question

I've been trying to use a regular expression to match and extract parts of a URL. The URL pattern looks like:

http://domain.abcdef/xyz/fe/fi/fo5/fu2m/123/

I intend to capture the following groups:

1. match and capture xyz (optional, but specific value)
2. match and capture fe/fi/fo5/fu2m (must exist, arbitrary value)
3. match and capture 123 (optional numeric value, which must appear at the end)

Here are expressions I have tried and problem encountered:

string1: http://domain.abcdef/xyz/fe/fi/fo5/fu2m/123/

string2: http://domain.abcdef/xyz/fe/fi/fo5/fu2m/

^(?:https?:\/\/)?(?:[\da-z\.-]+)\.(?:[a-z\.]{2,6})(?:\/(xyz))?\/([\/\w]+)+(?:\/([\d]+))\/$

• makes number at end mandatory

• matches and captures all groups as required in string1 even when xyz is not included

• no match in string2 because there's no number at the end

   ^(?:https?:\/\/)?(?:[\da-z\.-]+)\.(?:[a-z\.]{2,6})(?:\/(xyz))?\/([\/\w]+)+(?:\/([\d]+))?\/$
  

• makes number at end optional

• captures only groups 1 and 2 in string1 and string2 . Number is matched along with group 2 in string2 as fe/fi/fo5/fu2m/123

My problem is how to capture groups 1, 2 and 3 in all scenarios incl. string1 and string2 (note: I am using PHP's preg_match function)

Answer 1

I will use parse_url first to extract the path from the url. Then all you have to do is to use a non-greedy quantifier in the second group :

$path = parse_url($url, PHP_URL_PATH);

if ( preg_match('~^\A/([^/]+)/(.*?)/(?:(\d+)/)?\z~', $path, $m) )
    var_dump($m);

This way, if the number at the end is missing, the non-greedy quantifier (from the second group) is forced to reach the end of the string.