I have regex to parse all hash url in HTML content.
/(\#)([^\s]+")/g
HTML content will be as
Some text <a href="#some-hash1">some link</a>some content <a href="#some-hash2">some link1</a>
Expected is
#some-hash1, #some-hash2
But current regex is returning as (ending double come along with hash):
#some-hash1", #some-hash2"
I am unable to understand why its come along with double quotes. Any suggestion that will be very helpful.
I wouldn't use regex for this because it's overkill and because you can simply loop through the anchors pulling the value of their hrefs...
var anchors = document.querySelectorAll('a');
var hrefs = [];
anchors.forEach(function(e){
hrefs.push(e.getAttribute('href'));
});
console.log(hrefs);<a href="link 1">link 1</a>
<a href="link 2">link 2</a>Use non-capturing parenthesis,
/(\#)([^\s]+(?="))/g
DEMO
var z = 'Some text <a href="#some-hash1">some link</a>some content <a href="#some-hash2">some link1</a>';
console.log( z.match(/(\#)([^\s]+(?="))/g) );Just move double quote out the brackets:
(\#)([^\s]+)"
See how it works: https://regex101.com/r/fmrDyu/1
I am assuming that you are looking at the content of $2 for your result.
If so, the problem is the " inside the second capture group. Changing /(\#)([^\s]+")/g to /(\#)([^\s]+")/g results in the correct result.
I suggest joining the capture groups. Then /(\#[^\s]+)"/g will return $1=>#some-hash1, #some-hash2
Since $1 will always just return #, I suppose you trim it off elsewhere in your program, so perhaps you should use /\#([^\s]+)"/g which will return some-hash1, some-hash2 without the #
