I'm trying to detect the URI with a possibility of there being get params etc. I just need to get the /val1/val2 regardless if there is any url get params etc. how can I do this in php?
Thanks
$regex = '/^\/index.php?(.*)?\?+$/';
// Here are some example URLS
site.com/val1/val2
site.com/index.php/val1/val2
site.com/index.php/val1/val2?get1=val1&get2=val2
I want to be able to group the (/val1/val2) no matter what is on the left or right of the uri
You may use
(?:/[^/?#]+){2}(?=[?#]|$)
See the regex demo
Details
(?:/[^/?#]+){2} - two repetitions of / and then 1+ chars other than /, ? and #(?=[?#]|$) - a positive lookahead that matches a location immediately followed with ?, # or end of string.Here is the regex graph:
$re = '~(?:/[^/?#]+){2}(?=[?#]|$)~';
$str = 'site.com/val1/val2';
if (preg_match($re, $str, $match)) {
print_r($match[0]);
} // => /val1/val2
/, I am using ~ as a regex delimiter. You may see that no escaping is necessary here. If you handle the strings individually you may even replace (?=[?#]|$) with (?![^?#]) for better performance.
We can first try splitting the string on ?, to remove the query string, if it exists. Then, split by / path separator and access the last two paths:
$url = "site.com/index.php/val1/val2?get1=val1&get2=val2";
$url = explode('?', $url)[0];
$paths = explode('/', $url);
echo "second to last element: " . $paths[count($paths)-2] . "\n";
echo "last element: " . $paths[count($paths)-1];
This prints:
second to last element: val1
last element: val2
This answer avoids using regex entirely, which might result in slightly better performance.