忽略<a>[url][img]</a>内的所有内容的正则表达式

Question

This is a tricky regex code, actually i have a script that works fine, but i need a more precise pattern to avoid conflict with url that contain hashtag and bbcode.

actually my code replace all hashtags words like #example and turn it into link like:

https://myurl/?q=example

so my regex is:

const regex = /#[^\s.@#$%^&*()=+,\/;\[{\]}:?'";><]+/g;

i want a regex that ignore all urls inside <a> , all url inside [url], and [img] to avoid bbcode conflict.

but i can't find a valid pattern

full code that i have actually:

export default function () {

  const regex = /#[^\s!@#$%^&*()=+.\/,\[{\]};:'"?><]+/g;

  const p = this.$('.Post-body');
  const baseurl = app.forum.attribute('baseUrl');

  p.html = p.html(p.html().replace(regex, match => `<a href="${baseurl}/?q=${match}" class="hashlink" title="Find more post with this hashtag">${match}</a>`))

}

Answer 1

actually i've found a solution that works, i will post here the soution (could be useful for someone)

 const regex = /(?<!https?:\/\/\S*)#[^\s!@#$%^&*()=+.\/,\[{\]};:'"?><]+/g;

and i've removed the extra # using match.slice