preg_replace_callback occurrences of match unless a backslash exists 0-20 chars ahead

Question

For a long time I've been using preg_replace_callback to replace any occurrence of a reference ID into a link that opens the reference.

For example, "BIZ1234" would be replaced with Name title (pseudo code)

All good.

But now I need to not do the replacing if a backslash precedes the match.

For example, if the "BIZ1234" is used in a folder path: \drive_C\BIZ1234

For this I've tried a lot of regular expressions, and almost got what I needed, but my skills in regular expressions just aren't enough.

Current regular expression:

/BIZ(\d+)/i

Attempts of excluding occurrences where a backslash precedes the match:

/[^\\\\]BIZ(\d+)/i

This works if the backslash is right infront of BIZ. I realize I have to set some kind of limit as to how far back the backslash can be. So I try:

/[^\\\\](.{0,20})BIZ(\d+)/i

This replaces too much and doesn't care about the backslash.

I've read and read, but just can't grasp these regular expressions. Help! :-)

EDIT / ADDED:

I'm using the following as a test string (the following is 1 string):

<p>This should become a link BIZ1234</p>
<p>this/here/should/too BIZ1234</p>
<p>but\this\here\shouldn't BIZ1234</p>

Currently, using this regular expression:

/^([^\\\]+)BIZ(\d+)/

Replaces nothing in the above string.

Answer 1

As a lookbehind alternative, could use \K to reset the beginning of the match:

$pattern = '/>[^<\\\]*\KBIZ(\d+)/';

This would replace only from where you put the \K.

So for example match > followed by [^<\\\]* any amount of characters that are NOT < or backslash, followed by BIZ but reset and start replace just before BIZ.

Example at eval.in