php regex for detecting #number - php

Question

i have the following regex that i am trying to detect #x, x being a number. I was able to get it working when there is nothing around match 2, however if there is then it breaks. can someone help me with how to make this work both ways?
/(\G|\s+|^)#(\d+)((?=\s+)|(?=::)|$)/i
that will work with the line
This is a test #1234 end test
but that will not work with
This is a test #1234end test
This is a test#1234 end test
This is a test.#1234 end test
This is a test #1234. End test
anyone know what needs to be changed to achieve this?
edit, i am trying to allow anything but alphanumeric in the 3rd group, right now there is :: and whitespace. is there a way to combine these into 1 and not detect letters or numbers

Answer 1

Running a preg match using /#\d+/i should get you what you are looking for. So running the following:
$items = [
"This is a test #1234end test",
"This is a test#1234 end test",
"This is a test.#1234 end test",
"This is a test #1234. End test"
];
foreach($items as $test){
preg_match("/#\d+/i", $test, $matches);
var_dump($matches);
}
You will get this result:
array(1) {
[0]=>
string(5) "#1234"
}
array(1) {
[0]=>
string(5) "#1234"
}
array(1) {
[0]=>
string(5) "#1234"
}
array(1) {
[0]=>
string(5) "#1234"
}
If you don't want the # in the results, then you can then do a subpattern of /#(\d+)/i
Which will then result in the following:
array(2) {
[0]=>
string(5) "#1234"
[1]=>
string(4) "1234"
}
array(2) {
[0]=>
string(5) "#1234"
[1]=>
string(4) "1234"
}
array(2) {
[0]=>
string(5) "#1234"
[1]=>
string(4) "1234"
}
array(2) {
[0]=>
string(5) "#1234"
[1]=>
string(4) "1234"
}

Answer 2

(\G|\s+|^)#(\d+)((?=[^[:alnum:]])|$)
i wanted to keep the three groups that i had, but i only changed the 3rd group. i removed the :: and \S whitespace characters from the 3rd group and just added a simple NOT alphanumeric check, as this will contain those 2 conditions as well.
(\G|\s+|^)
#(\d+)
((?=[^[:alnum:]])|$)
[^[:alnum:]]