如何在行首_or_末尾匹配正则表达式？

Question

I'm trying to create a JS regex that matches !next either at the beginning or end of the line.

Currently I'm using /(^!next|!next$)/i which works. However that's obviously kind of ugly to read and not DRY, so I'm looking to improve.

The closest I've come is this:

/(?=!next)[^$]/i

But that actually just matches everything (and it occurs to me now that I misremembered the way that ^ works inside character classes, so this solution is garbage anyway).

I've searched the web and SO, and I'm totally stumped.

Answer 1

Here's a fun one, using regex conditionals:

/(^)?!next(?(1)|$)/gm

Demo on Regex101

How it works is it captures the beginning-of-line anchor, and, if it is not present, matches the end-of-line anchor instead.

Personally, though, I'd still argue in favour of your solution over mine because it's more readable to someone who doesn't have an extremely thorough knowledge of regexes (and is more portable).

For added fun, here's another version (that's even uglier than your initial variant):

/!next(?:(?<=^.{5})|(?=$))/gm

Demo on Regex101

I'd still recommend sticking with the classic alternation, though.

And, finally, one that works in JS (no, really):

/(?:^|(?=.{5}$))!next/gm

Demo on Regex101

Answer 2

You could try

function beginningOrEnd(input, search) {
  const pos = input.search(search);
  return pos === 0 || pos === input.length - search.length;
}

beginningOrEnd("hey, !help", "!help");

This uses search.length , so obviously search must be a string.

Or, you could construct the regexp yourself:

function beginningOrEnd(search) {
  return new RegExp(`^${search}|${search}\$`, "i");
}

beginningOrEnd("!help").test(input)

In the real world, you'd want to escape special regexp characters appropriately, of course.