[removed] negative lookbehind equivalent?

Question

Is there a way to achieve the equivalent of a negative lookbehind in javascript regular expressions? I need to match a string that does not start with a specific set of cha

Answer 1

following the idea of Mijoja, and drawing from the problems exposed by JasonS, i had this idea; i checked a bit but am not sure of myself, so a verification by someone more expert than me in js regex would be great :)

var re = /(?=(..|^.?)(ll))/g
         // matches empty string position
         // whenever this position is followed by
         // a string of length equal or inferior (in case of "^")
         // to "lookbehind" value
         // + actual value we would want to match

,   str = "Fall ball bill balll llama"

,   str_done = str
,   len_difference = 0
,   doer = function (where_in_str, to_replace)
    {
        str_done = str_done.slice(0, where_in_str + len_difference)
        +   "[match]"
        +   str_done.slice(where_in_str + len_difference + to_replace.length)

        len_difference = str_done.length - str.length
            /*  if str smaller:
                    len_difference will be positive
                else will be negative
            */

    }   /*  the actual function that would do whatever we want to do
            with the matches;
            this above is only an example from Jason's */



        /*  function input of .replace(),
            only there to test the value of $behind
            and if negative, call doer() with interesting parameters */
,   checker = function ($match, $behind, $after, $where, $str)
    {
        if ($behind !== "ba")
            doer
            (
                $where + $behind.length
            ,   $after
                /*  one will choose the interesting arguments
                    to give to the doer, it's only an example */
            )
        return $match // empty string anyhow, but well
    }
str.replace(re, checker)
console.log(str_done)

my personal output:

Fa[match] ball bi[match] bal[match] [match]ama

the principle is to call checker at each point in the string between any two characters, whenever that position is the starting point of:

--- any substring of the size of what is not wanted (here 'ba', thus ..) (if that size is known; otherwise it must be harder to do perhaps)

--- --- or smaller than that if it's the beginning of the string: ^.?

and, following this,

--- what is to be actually sought (here 'll').

At each call of checker, there will be a test to check if the value before ll is not what we don't want (!== 'ba'); if that's the case, we call another function, and it will have to be this one (doer) that will make the changes on str, if the purpose is this one, or more generically, that will get in input the necessary data to manually process the results of the scanning of str.

here we change the string so we needed to keep a trace of the difference of length in order to offset the locations given by replace, all calculated on str, which itself never changes.

since primitive strings are immutable, we could have used the variable str to store the result of the whole operation, but i thought the example, already complicated by the replacings, would be clearer with another variable (str_done).

i guess that in terms of performances it must be pretty harsh: all those pointless replacements of '' into '', this str.length-1 times, plus here manual replacement by doer, which means a lot of slicing... probably in this specific above case that could be grouped, by cutting the string only once into pieces around where we want to insert [match] and .join()ing it with [match] itself.

the other thing is that i don't know how it would handle more complex cases, that is, complex values for the fake lookbehind... the length being perhaps the most problematic data to get.

and, in checker, in case of multiple possibilities of nonwanted values for $behind, we'll have to make a test on it with yet another regex (to be cached (created) outside checker is best, to avoid the same regex object to be created at each call for checker) to know whether or not it is what we seek to avoid.

hope i've been clear; if not don't hesitate, i'll try better. :)

Answer 2

Using your case, if you want to replace m with something, e.g. convert it to uppercase M, you can negate set in capturing group.

match ([^a-g])m, replace with $1M

"jim jam".replace(/([^a-g])m/g, "$1M")
\\jiM jam

([^a-g]) will match any char not(^) in a-g range, and store it in first capturing group, so you can access it with $1.

So we find im in jim and replace it with iM which results in jiM.

Answer 3

You could define a non-capturing group by negating your character set:

(?:[^a-g])m

...which would match every m NOT preceded by any of those letters.

Answer 4

This is how I achieved str.split(/(?<!^)@/) for Node.js 8 (which doesn't support lookbehind):

str.split('').reverse().join('').split(/@(?!$)/).map(s => s.split('').reverse().join('')).reverse()

Works? Yes (unicode untested). Unpleasant? Yes.

Answer 5

This effectively does it

"jim".match(/[^a-g]m/)
> ["im"]
"jam".match(/[^a-g]m/)
> null

Search and replace example

"jim jam".replace(/([^a-g])m/g, "$1M")
> "jiM jam"

Note that the negative look-behind string must be 1 character long for this to work.

Answer 6

As mentioned before, JavaScript allows lookbehinds now. In older browsers you still need a workaround.

I bet my head there is no way to find a regex without lookbehind that delivers the result exactly. All you can do is working with groups. Suppose you have a regex (?<!Before)Wanted, where Wanted is the regex you want to match and Before is the regex that counts out what should not precede the match. The best you can do is negate the regex Before and use the regex NotBefore(Wanted). The desired result is the first group $1.

In your case Before=[abcdefg] which is easy to negate NotBefore=[^abcdefg]. So the regex would be [^abcdefg](m). If you need the position of Wanted, you must group NotBefore too, so that the desired result is the second group.

If matches of the Before pattern have a fixed length n, that is, if the pattern contains no repetitive tokens, you can avoid negating the Before pattern and use the regular expression (?!Before).{n}(Wanted), but still have to use the first group or use the regular expression (?!Before)(.{n})(Wanted) and use the second group. In this example, the pattern Before actually has a fixed length, namely 1, so use the regex (?![abcdefg]).(m) or (?![abcdefg])(.)(m). If you are interested in all matches, add the g flag, see my code snippet:

function TestSORegEx() {
  var s = "Donald Trump doesn't like jam, but Homer Simpson does.";
  var reg = /(?![abcdefg])(.{1})(m)/gm;
  var out = "Matches and groups of the regex " + 
            "/(?![abcdefg])(.{1})(m)/gm in \ns = \"" + s + "\"";
  var match = reg.exec(s);
  while(match) {
    var start = match.index + match[1].length;
    out += "\nWhole match: " + match[0] + ", starts at: " + match.index
        +  ". Desired match: " + match[2] + ", starts at: " + start + ".";   
    match = reg.exec(s);
  }
  out += "\nResulting string after statement s.replace(reg, \"$1*$2*\")\n"
         + s.replace(reg, "$1*$2*");
  alert(out);
}