How to match !x but not !!x in JS regex?

Question

Given the following text:

This is!!xa simple string!xpattern

I would like to get a regexp that matches the !x that's between "string" and "pattern" but not !!xa that's between "is" and "a".

This regexp is to be used inside a string split().

I have tried several combinations but I cannot get a regexp that meets my needs. Perhaps my expression is not so regular after all =)

Thanks in advance!

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EDIT:

SOLUTION

Just to state clear the solution is going to be:

s.replace(/(([^!])|^)!x/g,'$1SOME_MAGICAL_STRING').split(/SOME_MAGICAL_STRING/)

Thanks for the solution idea to both jvenema and Amarghosh. And also to everyone that provided feedback too.

Answer 1

This expression should do the job. It uses a negative look behind assertion to assert that there is only a single exclamation mark.

(?<!!)!x

Answer 2

var s = "This is!!xa simple string!xpattern";
s.replace(/[^!]!x/,'-');

output:

"This is!!xa simple strin-pattern"

Edit: I missed the g, my bad. This one works:

var s = "!xThis is!!xa simple string!xpattern";
s.replace(/(([^!])|^)!x/g,'$1-');

output:

"-This is!!xa simple string-pattern"

All we're doing is matching the preceding character and then including it back into the replacement.

Answer 3

Too bad JS doesn't have lookbehind :) Assuming no !x!x, You can use RegExp.exec instead of String.split, as in

rx = /((?:[^!]|![^x])+)(?:!x|$)/g
res = []
while ((m = rx.exec("This is!!xa simple string!xpattern")))
  res.push(m[1]);

Here, (?:[^!]|![^x])+ matches one or more non-exclamation point, or a ! not followed by an x. This latter case gets rid of the !!. The (?:!x|$) consumes the !x terminator.

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Edit: Since !x!x can happen, the loop has to be modified a bit to avoid infinite loop.

rx = /((?:[^!]|![^x])*)(?:!x|$)/g
res = []
str = "This is!!xa simpl!!xe!x!x string!xpattern"
while (true) {
  var m = rx.exec(str);
  if (m.index >= str.length)
    break;
  res.push(m[1]);
}
res