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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 characters.
It seems I am unable to find a regex that does this without failing if the matched part is found at the beginning of the string. Negative lookbehinds seem to be the only answer, but JavaScript doesn't has one.
This is the regex that I would like to work, but it doesn't:
(?<!([abcdefg]))m
So it would match the 'm' in 'jim' or 'm', but not 'jam'
Since 2018, Lookbehind Assertions are part of the ECMAScript language specification.
// positive lookbehind
(?<=...)
// negative lookbehind
(?<!...)
Answer pre-2018
As Javascript supports negative lookahead, one way to do it is:
reverse the input string
match with a reversed regex
reverse and reformat the matches
const reverse = s => s.split('').reverse().join('');
const test = (stringToTests, reversedRegexp) => stringToTests
.map(reverse)
.forEach((s,i) => {
const match = reversedRegexp.test(s);
console.log(stringToTests[i], match, 'token:', match ? reverse(reversedRegexp.exec(s)[0]) : 'Ø');
});
Example 1:
Following #andrew-ensley's question:
test(['jim', 'm', 'jam'], /m(?!([abcdefg]))/)
Outputs:
jim true token: m
m true token: m
jam false token: Ø
Example 2:
Following #neaumusic comment (match max-height but not line-height, the token being height):
test(['max-height', 'line-height'], /thgieh(?!(-enil))/)
Outputs:
max-height true token: height
line-height false token: Ø
Lookbehind Assertions got accepted into the ECMAScript specification in 2018.
Positive lookbehind usage:
console.log(
"$9.99 €8.47".match(/(?<=\$)\d+\.\d*/) // Matches "9.99"
);
Negative lookbehind usage:
console.log(
"$9.99 €8.47".match(/(?<!\$)\d+\.\d*/) // Matches "8.47"
);
Platform support:
✔️ V8
✔️ Google Chrome 62.0
✔️ Microsoft Edge 79.0
✔️ Node.js 6.0 behind a flag and 9.0 without a flag
✔️ Deno (all versions)
✔️ SpiderMonkey
✔️ Mozilla Firefox 78.0
🛠️ JavaScriptCore: support in beta as of 2023-02-16: the feature has been merged
✔️ Apple Safari 16.4 (in beta as of 2023-02-16)
✔️ iOS 16.4 beta WebView (all browsers on iOS + iPadOS)
✔️ Bun 0.2.2
❌ Chakra: Microsoft was working on it but Chakra is now abandoned in favor of V8
❌ Internet Explorer
❌ Edge versions prior to 79 (the ones based on EdgeHTML+Chakra)
Let's suppose you want to find all int not preceded by unsigned :
With support for negative look-behind:
(?<!unsigned )int
Without support for negative look-behind:
((?!unsigned ).{9}|^.{0,8})int
Basically idea is to grab n preceding characters and exclude match with negative look-ahead, but also match the cases where there's no preceeding n characters. (where n is length of look-behind).
So the regex in question:
(?<!([abcdefg]))m
would translate to:
((?!([abcdefg])).|^)m
You might need to play with capturing groups to find exact spot of the string that interests you or you want to replace specific part with something else.
Mijoja's strategy works for your specific case but not in general:
js>newString = "Fall ball bill balll llama".replace(/(ba)?ll/g,
function($0,$1){ return $1?$0:"[match]";});
Fa[match] ball bi[match] balll [match]ama
Here's an example where the goal is to match a double-l but not if it is preceded by "ba". Note the word "balll" -- true lookbehind should have suppressed the first 2 l's but matched the 2nd pair. But by matching the first 2 l's and then ignoring that match as a false positive, the regexp engine proceeds from the end of that match, and ignores any characters within the false positive.
Use
newString = string.replace(/([abcdefg])?m/, function($0,$1){ return $1?$0:'m';});
You could define a non-capturing group by negating your character set:
(?:[^a-g])m
...which would match every m NOT preceded by any of those letters.
This is how I achieved str.split(/(?<!^)#/) for Node.js 8 (which doesn't support lookbehind):
str.split('').reverse().join('').split(/#(?!$)/).map(s => s.split('').reverse().join('')).reverse()
Works? Yes (unicode untested). Unpleasant? Yes.
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. :)
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.
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);
}
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.
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 characters.
It seems I am unable to find a regex that does this without failing if the matched part is found at the beginning of the string. Negative lookbehinds seem to be the only answer, but JavaScript doesn't has one.
This is the regex that I would like to work, but it doesn't:
(?<!([abcdefg]))m
So it would match the 'm' in 'jim' or 'm', but not 'jam'
Since 2018, Lookbehind Assertions are part of the ECMAScript language specification.
// positive lookbehind
(?<=...)
// negative lookbehind
(?<!...)
Answer pre-2018
As Javascript supports negative lookahead, one way to do it is:
reverse the input string
match with a reversed regex
reverse and reformat the matches
const reverse = s => s.split('').reverse().join('');
const test = (stringToTests, reversedRegexp) => stringToTests
.map(reverse)
.forEach((s,i) => {
const match = reversedRegexp.test(s);
console.log(stringToTests[i], match, 'token:', match ? reverse(reversedRegexp.exec(s)[0]) : 'Ø');
});
Example 1:
Following #andrew-ensley's question:
test(['jim', 'm', 'jam'], /m(?!([abcdefg]))/)
Outputs:
jim true token: m
m true token: m
jam false token: Ø
Example 2:
Following #neaumusic comment (match max-height but not line-height, the token being height):
test(['max-height', 'line-height'], /thgieh(?!(-enil))/)
Outputs:
max-height true token: height
line-height false token: Ø
Lookbehind Assertions got accepted into the ECMAScript specification in 2018.
Positive lookbehind usage:
console.log(
"$9.99 €8.47".match(/(?<=\$)\d+\.\d*/) // Matches "9.99"
);
Negative lookbehind usage:
console.log(
"$9.99 €8.47".match(/(?<!\$)\d+\.\d*/) // Matches "8.47"
);
Platform support:
✔️ V8
✔️ Google Chrome 62.0
✔️ Microsoft Edge 79.0
✔️ Node.js 6.0 behind a flag and 9.0 without a flag
✔️ Deno (all versions)
✔️ SpiderMonkey
✔️ Mozilla Firefox 78.0
🛠️ JavaScriptCore: support in beta as of 2023-02-16: the feature has been merged
✔️ Apple Safari 16.4 (in beta as of 2023-02-16)
✔️ iOS 16.4 beta WebView (all browsers on iOS + iPadOS)
✔️ Bun 0.2.2
❌ Chakra: Microsoft was working on it but Chakra is now abandoned in favor of V8
❌ Internet Explorer
❌ Edge versions prior to 79 (the ones based on EdgeHTML+Chakra)
Let's suppose you want to find all int not preceded by unsigned :
With support for negative look-behind:
(?<!unsigned )int
Without support for negative look-behind:
((?!unsigned ).{9}|^.{0,8})int
Basically idea is to grab n preceding characters and exclude match with negative look-ahead, but also match the cases where there's no preceeding n characters. (where n is length of look-behind).
So the regex in question:
(?<!([abcdefg]))m
would translate to:
((?!([abcdefg])).|^)m
You might need to play with capturing groups to find exact spot of the string that interests you or you want to replace specific part with something else.
Mijoja's strategy works for your specific case but not in general:
js>newString = "Fall ball bill balll llama".replace(/(ba)?ll/g,
function($0,$1){ return $1?$0:"[match]";});
Fa[match] ball bi[match] balll [match]ama
Here's an example where the goal is to match a double-l but not if it is preceded by "ba". Note the word "balll" -- true lookbehind should have suppressed the first 2 l's but matched the 2nd pair. But by matching the first 2 l's and then ignoring that match as a false positive, the regexp engine proceeds from the end of that match, and ignores any characters within the false positive.
Use
newString = string.replace(/([abcdefg])?m/, function($0,$1){ return $1?$0:'m';});
You could define a non-capturing group by negating your character set:
(?:[^a-g])m
...which would match every m NOT preceded by any of those letters.
This is how I achieved str.split(/(?<!^)#/) for Node.js 8 (which doesn't support lookbehind):
str.split('').reverse().join('').split(/#(?!$)/).map(s => s.split('').reverse().join('')).reverse()
Works? Yes (unicode untested). Unpleasant? Yes.
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. :)
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.
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);
}
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, "$1M")
> "jiM jam"
Note that the negative look-behind string must be 1 character long for this to work.
So I've got this RegExp to validate some input like this:
1 12919840 T C
1 35332717 C A
1 55148456 G T
1 70504789 C T
1 167059520 A T
1 182496864 A T
1 197073351 C T
1 216373211 G T
The exp i came up with is:
/^([0-9]\s+[0-9]+\s+[ATCG]\s+[ATCG][\s|\n]+)*[0-9]\s+[0-9]+\s+[ATCG]\s+[ATCG][\s|\n]*$/g
This worked in something like
/^([0-9]\s+[0-9]+\s+[ATCG]\s+[ATCG][\s|\n]+)*[0-9]\s+[0-9]+\s+[ATCG]\s+[ATCG][\s|\n]*$/g.test("1 12919840 T C\n1 35332717 C A"); //this returns true
But when trying use group reference to make it shorter it doesn't work anymore
/^(([0-9]\s+[0-9]+\s+[ATCG]\s+[ATCG])[\s|\n]+)*\2[\s|\n]*$/g.test("1 12919840 T C\n1 35332717 C A"); //this returns false
I'm using \2 here since from my research the numbering of the groups starts from the left most parenthesis. what did I miss? thx!
My answer addresses your what did I miss? question. For a workaround, see the answer by #Jack.
A Capture Group is Not a Subroutine
What you're missing is that a capture group is not a subroutine.
When you say \1, you are referring to the exact characters that were captured by the parentheses of Group 1. For instance, (\d)\1 would match 11 or 22, but not 12.
Regex Subroutines
In Perl and PCRE, you can refer to a subexpression by using (?1). For instance, (\d)(?1) would match 11 as well as 12.
This is also available in the regex module for Python. Sadly, this is not available in JavaScript, which you seem to be using.
Since you're working with DNA, if you have a chance, I'd suggest working in a life-embracing language such as Python (JS has poor regex abilities, though the XregeExp library fills some holes.)
The problem with your expression is that back references match whatever was matched by a previous memory capture; the expression that generated the memory capture itself can't be referenced in this manner.
That said, you can still shorten the expression by using a multiplier on your expression:
var re = /((?:^\d+\s+\d+\s+[TCGA]\s+[TCGA][\s\r\n]*){2})/gm,
m;
The expression matches the same thing twice, with an optional set of spaces in between. To iterate over all matches:
while ((m = re.exec(str)) !== null) {
console.log('match' + m[1]);
}
I've spent a few hours on this and I can't seem to figure this one out.
In the code below, I'm trying to understand exactly what and how the regular expressions in the url.match are working.
As the code is below, it doesn't work. However if I remove (?:&toggle=|&ie=utf-8|&FORM=|&aq=|&x=|&gwp) it seems to give me the output that I want.
However, I don't want to remove this without understanding what it is doing.
I found a pretty useful resource, but after a few hours I still can't precisely determine what these expressions are doing:
https://developer.mozilla.org/en-US/docs/JavaScript/Guide/Regular_Expressions#Using_Parenthesized_Substring_Matches
Could someone break this down for me and explain how exactly it is parsing the strings. The expressions themselves and the placement of the parentheses is not really clear to me and frankly very confusing.
Any help is appreciated.
(function($) {
$(document).ready(function() {
function parse_keywords(url){
var matches = url.match(/.*(?:\?p=|\?q=|&q=|\?s=)([a-zA-Z0-9 +]*)(?:&toggle=|&ie=utf-8|&FORM=|&aq=|&x=|&gwp)/);
return matches ? matches[1].split('+') : [];
}
myRefUrl = "http://www.google.com/url?sa=f&rct=j&url=https://www.mydomain.com/&q=my+keyword+from+google&ei=fUpnUaage8niAKeiICgCA&usg=AFQjCNFAlKg_w5pZzrhwopwgD12c_8z_23Q";
myk1 = (parse_keywords(myRefUrl));
kw="";
for (i=0;i<myk1.length;i++) {
if (i == (myk1.length - 1)) {
kw = kw + myk1[i];
}
else {
kw = kw + myk1[i] + '%20';
}
}
console.log (kw);
if (kw != null && kw != "" && kw != " " && kw != "%20") {
orighref = $('a#applynlink').attr('href');
$('a#applynlink').attr('href', orighref + '&scbi=' + kw);
}
});
})(jQuery);
Let's break this regex down.
/
Begin regex.
.*
Match zero or more anything - basically, we're willing to match this regex at any point into the string.
(?:\?p=
|\?q=
|&q=
|\?s=)
In this, the ?: means 'do not capture anything inside of this group'. See http://www.regular-expressions.info/refadv.html
The \? means take ? literally, which is normally a character meaning 'match 0 or 1 copies of the previous token' but we want to match an actual ?.
Other than that, it's just looking for a multitude of different options to select (| means 'the regex is valid if I match either what's before me or after me.)
([a-zA-Z0-9 +]*)
Now we match zero or more of any of the following characters in any arrangement: a-ZA-Z0-9 + And since it is inside a () with no ?: we DO capture it.
(?:&toggle=
|&ie=utf-8
|&FORM=
|&aq=
|&x=
|&gwp)
We see another ?: so this is another non-capturing group.
Other than that, it is just full of literal characters separated by |s, so it is not doing any fancy logic.
/
End regex.
In summary, this regex looks through the string for any instance of the first non capturing group, captures everything inside of it, then looks for any instance of the second non capturing group to 'cap' it off and returns everything that was between those two non capturing groups. (Think of it as a 'sandwich', we look for the header and footer and capture everything in between that we're interested in)
After the regex runs, we do this:
return matches ? matches[1].split('+') : [];
Which grabs the captured group and splits it on + into an array of strings.
For situations like this, it's really helpful to visualize it with www.debuggex.com (which I built). It immediately shows you the structure of your regex and allows you to walk through step-by-step.
In this case, the reason it works when you remove the last part of your regex is because none of the strings &toggle=, &ie=utf-8, etc are in your sample url. To see this, drag the grey slider above the test string on debuggex and you'll see that it never makes it past the & in that last group.
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 characters.
It seems I am unable to find a regex that does this without failing if the matched part is found at the beginning of the string. Negative lookbehinds seem to be the only answer, but JavaScript doesn't has one.
This is the regex that I would like to work, but it doesn't:
(?<!([abcdefg]))m
So it would match the 'm' in 'jim' or 'm', but not 'jam'
Since 2018, Lookbehind Assertions are part of the ECMAScript language specification.
// positive lookbehind
(?<=...)
// negative lookbehind
(?<!...)
Answer pre-2018
As Javascript supports negative lookahead, one way to do it is:
reverse the input string
match with a reversed regex
reverse and reformat the matches
const reverse = s => s.split('').reverse().join('');
const test = (stringToTests, reversedRegexp) => stringToTests
.map(reverse)
.forEach((s,i) => {
const match = reversedRegexp.test(s);
console.log(stringToTests[i], match, 'token:', match ? reverse(reversedRegexp.exec(s)[0]) : 'Ø');
});
Example 1:
Following #andrew-ensley's question:
test(['jim', 'm', 'jam'], /m(?!([abcdefg]))/)
Outputs:
jim true token: m
m true token: m
jam false token: Ø
Example 2:
Following #neaumusic comment (match max-height but not line-height, the token being height):
test(['max-height', 'line-height'], /thgieh(?!(-enil))/)
Outputs:
max-height true token: height
line-height false token: Ø
Lookbehind Assertions got accepted into the ECMAScript specification in 2018.
Positive lookbehind usage:
console.log(
"$9.99 €8.47".match(/(?<=\$)\d+\.\d*/) // Matches "9.99"
);
Negative lookbehind usage:
console.log(
"$9.99 €8.47".match(/(?<!\$)\d+\.\d*/) // Matches "8.47"
);
Platform support:
✔️ V8
✔️ Google Chrome 62.0
✔️ Microsoft Edge 79.0
✔️ Node.js 6.0 behind a flag and 9.0 without a flag
✔️ Deno (all versions)
✔️ SpiderMonkey
✔️ Mozilla Firefox 78.0
🛠️ JavaScriptCore: support in beta as of 2023-02-16: the feature has been merged
✔️ Apple Safari 16.4 (in beta as of 2023-02-16)
✔️ iOS 16.4 beta WebView (all browsers on iOS + iPadOS)
✔️ Bun 0.2.2
❌ Chakra: Microsoft was working on it but Chakra is now abandoned in favor of V8
❌ Internet Explorer
❌ Edge versions prior to 79 (the ones based on EdgeHTML+Chakra)
Let's suppose you want to find all int not preceded by unsigned :
With support for negative look-behind:
(?<!unsigned )int
Without support for negative look-behind:
((?!unsigned ).{9}|^.{0,8})int
Basically idea is to grab n preceding characters and exclude match with negative look-ahead, but also match the cases where there's no preceeding n characters. (where n is length of look-behind).
So the regex in question:
(?<!([abcdefg]))m
would translate to:
((?!([abcdefg])).|^)m
You might need to play with capturing groups to find exact spot of the string that interests you or you want to replace specific part with something else.
Mijoja's strategy works for your specific case but not in general:
js>newString = "Fall ball bill balll llama".replace(/(ba)?ll/g,
function($0,$1){ return $1?$0:"[match]";});
Fa[match] ball bi[match] balll [match]ama
Here's an example where the goal is to match a double-l but not if it is preceded by "ba". Note the word "balll" -- true lookbehind should have suppressed the first 2 l's but matched the 2nd pair. But by matching the first 2 l's and then ignoring that match as a false positive, the regexp engine proceeds from the end of that match, and ignores any characters within the false positive.
Use
newString = string.replace(/([abcdefg])?m/, function($0,$1){ return $1?$0:'m';});
You could define a non-capturing group by negating your character set:
(?:[^a-g])m
...which would match every m NOT preceded by any of those letters.
This is how I achieved str.split(/(?<!^)#/) for Node.js 8 (which doesn't support lookbehind):
str.split('').reverse().join('').split(/#(?!$)/).map(s => s.split('').reverse().join('')).reverse()
Works? Yes (unicode untested). Unpleasant? Yes.
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. :)
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.
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);
}
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.