Question

I’m trying to write a little WordPress plugin to support some migrated content.

The syntax highlighter expects (for proper highlighting):

<pre lang='something'>
  <code>
    The code...
  </code>
</pre>

However, my markdown code has the following:

<pre>
  <code>
    :::something
    The code...
  </code>
</pre>

I think you can see where this is going. What I want to achieve is this:

1. :::something should be removed, and the <pre> tag should be updated to <pre lang="something">.
2. If :::something does not exist, the <pre> tag should be <pre lang="plain">
3. There may be multiple occurrences per page that need to be updated.

How would a PHP function achieving the above look like?

function set_syntax_lang($content) {
  // Do stuff here
  return $new_content;
}

What I gathered so far is this regex:

/<pre.*>\s*<code>\s*:::(\w)/

This even yields me, using preg_match, the actual syntax indicator (something), but I don’t know how to update the pre-tag correctly.

It’s been a very long time since I coded PHP and regexes are not really my strong suit. So all help is appreciated.

Answer 1

Finding :::something

preg_replace( '/<pre(.*>\s*<code>\s*):::(\w+)/U', '<pre lang="$2"$1' , $html );

This is an edge-case. But normally I should advise you to NOT use regex for html (bobince someone?).

Also next time try be less verbouse on your question. I took more time to read you than to write this answer.

Finding code without :::something

preg_replace( '/<pre(.*>\s*<code>\s*)(?!:::\w+)/U', '<pre lang="plain"$1' , $html );

Fixing <code>

preg_replace( array( '/(<pre.*>)\s*<code>/U' , '/<\/code>\s*(<\/pre>)/U' ),
              '$1' , $html );
//> Completly untested