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Regex replace not removing characters properly

I have the regular expression:
const regex = /^\d*\.?\d{0,2}$/
and its inverse (I believe) of
const inverse = /^(?!\d*\.?\d{0,2}$)/
The first regex is validating the string fits any positive number, allowing a decimal and two decimal digits (e.g. 150, 14., 7.4, 12.68). The second regex is the inverse of the first, and doing some testing I'm fairly confident it's giving the expected result, as it only validates when the string is anything but a number that may have a decimal and two digits after (e.g. 12..05, a5, 54.357).
My goal is to remove any characters from the string that do not fit the first regex. I thought I could do that this way:
let myString = '123M.45';
let fixed = myString.replace(inverse, '');
But this does not work as intended. To debug, I tried having the replace character changed to something I would be able to see:
let fixed = myString.replace(inverse, 'ZZZ');
When I do this, fixed becomes: ZZZ123M.45
Any help would be greatly appreciated.

I think I understand your logic here trying to find a regex that is the inverse of the regex that matches your valid string, in the hopes that it will allow you to remove any characters that make your string invalid and leave only the valid string. However, I don't think replace() will allow you to solve your problem in this way. From the MDN docs:
The replace() method returns a new string with some or all matches of a pattern replaced by a replacement.
In your inverse pattern you are using a negative lookahead. If we take a simple example of X(?!Y) we can think of this as "match X if not followed by Y". In your pattern your "X" is ^ and your "Y" is \d*\.?\d{0,2}$. From my understanding, the reason you are getting ZZZ123M.45 is that it is finding the first ^ (i.e, the start of the string) that is not followed by your pattern \d*\.?\d{0,2}$, and since 123M.45 doesn't match your "Y" pattern, your negative lookahead is satisfied and the beginning of your string is matched and "replaced" with ZZZ.
That (I think) is an explanation of what you are seeing.
I would propose an alternative solution to your problem that better fits with how I understand the .replace() method. Instead of your inverse pattern, try this one:
const invalidChars = /[^\d\.]|\.(?=\.)|(?<=\.\d\d)\d*/g
const myString = '123M..456444';
const fixed = myString.replace(invalidChars, '');
Here I am using a pattern that I think will match the individual characters that you want to remove. Let's break down what this one is doing:
[^\d\.]: match characters that are not digits
\.(?=\.): match . character if it is followed by another . character.
(?<=\.\d\d)\d*: match digits that are preceded by a decimal and 2 digits
Then I join all these with ORs (|) so it will match any one of the above patterns, and I use the g flag so that it will replace all the matches, not just the first one.
I am not sure if this will cover all your use cases, but I thought I would give it a shot. Here's a link to a breakdown that might be more helpful than mine, and you can use this tool to tweak the pattern if necessary.

I don't think you can do this
remove any characters from the string that do not fit the first regex
Because regex matching is meant for the entire string, and replace is used to replace just a PART inside that string. So the Regex inside replace must be a Regex to match unwanted characters only, not inverted Regex.
What you could do is to validate the string with your original regex, then if it's not valid, replace and validate again.
//if (notValid), replace unwanted character
// replace everything that's not a dot or digit
const replaceRegex = /[^\d.]/g; // notice g flag here to match every occurrence
const myString = '123M.45';
const fixed = myString.replace(replaceRegex, '');
console.log(fixed)
// validate again

Splitting a string into words and keeping delimiter

I want to split up a string (sentence) in an array of words and keep the delimiters.
I have found and I am currently using this regex for this:
[^.!?\s][^.!?]*(?:[.!?](?!['"]?\s|$)[^.!?]*)*[.!?]?['"]?(?=\s|$)
An explanation can be found here: http://regex101.com/
This works exactly as I want it to and effectively makes a string like
This is a sentence.
To an array of
["This", "is", "a", "sentence."]
The problem here is that it does not include spaces nor newlines. I want the string to be parsed as words as it already does but I also want the corresponding space and or newline character to belong to the previous word.
I have read about positive lookahead that should look for future characters (space and or newline) but still take them into account when extracting the word. Although this might be the solution I have failed to implement it.
If it makes any difference I am using JavaScript and the following code:
//save the regex -- g modifier to get all matches
var reg = /[^.!?\s][^.!?]*(?:[.!?](?!['"]?\s|$)[^.!?]*)*[.!?]?['"]?(?=\s|$)/g;
//define variable for holding matches
var matches;
//loop through each match
while(matches = reg.exec(STRING_HERE)){
//the word without spaces or newlines
console.log(matches[0]);
}
The code works but as I said, it does not include spaces and newline characters.

Yo can try something simpler:
str.split(/\b(?!\s)/);
However, note non word characters (e.g. full stop) will be considered another word:
"This is a sentence.".split(/\b(?!\s)/);
// [ "This ", "is ", "a ", "sentence", "." ]
To fix that, you can use a character class with the characters that shouldn't begin another word:
str.split(/\b(?![\s.])/);

function split_string(str){
var arr = str.split(" ");
var last_i = arr.length - 1;
for(var i=0; i<last_i; i++){
arr[i]+=" ";
}
return arr;
}

It may be as simple as this:
var sentence = 'This is a sentence.';
sentence = sentence.split(' ').join(' ||');
sentence = sentence.split('\n').join('\n||');
var matches = sentence.split('||');
Note that I use 2 pipes as a delimiter, but ofcourse you can use anything as long as it's unique.
Also note that I only split \n as a newline, but you may add \r\n or whatever you want to split as well.

General Solution
To keep the delimiters conjoined in the results, the regex needs to be a zero-width match. In other words, the regex can be thought of as matching the point between a delimiter and non-delimiter, rather than matching the delimiters themselves. This can be achieved with zero-width matching expressions, matching before, at, or after the split point (at most one each); let's call these A, B, and C. Sometimes a single sub-expression will do it, others you'll need two; offhand, I can't think of a case where you'd need three.
Not only look-aheads but lookarounds in general are the perfect candidates for this purpose: lookbehinds ((?<=...)) to match before the split point, and lookaheads ((?=...)) after. That's the essence of this approach. Positive or negative lookarounds can be used. The one pitfall is that lookbehinds are relatively new to JS regexes, so not all browsers or other JS engines will support them (current versions of Firefox, Chrome, Opera, Edge, and node.js do; Safari does not). If you need to support a JS engine that doesn't support lookbehinds, you might still be able to write & use a regex that matches at-and-before (BC).
To have the delimiters appear at the end of each match, put them in A. To have them at the start, in C. Fortunately, JS regexes do not place restrictions on lookbehinds, so simply wrapping the delimiter regex in the positive lookaround markers should be all that's required for delimiters. If the delimiters aren't so simple (i.e. context-sensitive), it might take a little more work to write the regex, which doesn't need to match the entire delimiter.
Paired with the delimiter pattern, you'll need to write a pattern that matches the start (for C) or end (for A) of the non-delimiter. This step is likely the one that will require the most additional work.
The at-split-point match, B
will often (always?) be a simple boundary, such as \b.
Specific Solution
If spaces are the only delimiters, and they're to appear at the end of each match, the delimiter pattern would be (?<=\s), in A. However, there are some cases not covered in the problem description. For example, should words separated by only punctuation (e.g. "x.y") be split? Which side of a split point should quotation marks and hyphens appear, if any? Should they count as punctuation? Another option for the delimiter is to match (after) all non-word characters, in which case A would be (<?=\W).
Since the split-point is at a word boundary, B could be \b.
Since the start of a match is a word character, (?=\w) will suffice for C.
Any two of those three should suffice. One that is perhaps clearest in meaning (and splits at the most points) is /(<?=\W)(?=\w)/, which can be translated as "split at the start of each word". \b could be added, if you find it more understandable, though it has no functional affect: /(<?=\W)\b(?=\w)/.
Note Oriol's excellent solutions are given by B=\b and (C=(?!\s) or C=(?![\s.])).
Additional
As a point of interest, there would be a simpler solution for this particular case if JS regexes supported TCL word boundaries: \m matches only at the start of a word, so str.split(/\m/) would split exactly at the start of each word. (\m is equivalent to (<?=\W)(?=\w).)

If you want to include the whitespace after the word, the regex \S+\s* should work.
const s = `This is a sentence.
This is another sentence.`;
console.log(s.match(/\S+\s*/g))

Regex to remove non-letter characters but keep accented letters

I have strings in Spanish and other languages that may contain generic special characters like (),*, etc. That I need to remove. But the problem is that it also may contain special language characters like ñ, á, ó, í etc and they need to remain. So I am trying to do it with regexp the following way:
var desired = stringToReplace.replace(/[^\w\s]/gi, '');
Unfortunately it is removing all special characters including the language related. Not sure how to avoid that. Maybe someone could suggest?

I would suggest using Steven Levithan's excellent XRegExp library and its Unicode plug-in.
Here's an example that strips non-Latin word characters from a string: http://jsfiddle.net/b3awZ/1/
var regex = XRegExp("[^\\s\\p{Latin}]+", "g");
var str = "¿Me puedes decir la contraseña de la Wi-Fi?"
var replaced = XRegExp.replace(str, regex, "");
See also this answer by Steven Levithan himself:
Regular expression Spanish and Arabic words

Instead of whitelisting characters you accept, you could try blacklisting illegal characters:
var desired = stringToReplace.replace(/[-'`~!##$%^&*()_|+=?;:'",.<>\{\}\[\]\\\/]/gi, '')

Note! Works only for 16bit code points. This answer is incomplete.
Short answer
The character class for all arabic digits and latin letters is: [0-9A-Za-z\u00c0-\u00d6\u00d8-\u00f6\u00f8-\u02af\u1d00-\u1d25\u1d62-\u1d65\u1d6b-\u1d77\u1d79-\u1d9a\u1e00-\u1eff\u2090-\u2094\u2184-\u2184\u2488-\u2490\u271d-\u271d\u2c60-\u2c7c\u2c7e-\u2c7f\ua722-\ua76f\ua771-\ua787\ua78b-\ua78c\ua7fb-\ua7ff\ufb00-\ufb06].
To get a regex you can use, prepend /^ and append +$/. This will match strings consisting of only latin letters and digits like "mérito" or "Schönheit".
To match non-digits or non-letter characters to remove them, write a ^ as first character after the opening bracket [ and prepend / and append +/.
How did I find that out? Continue reading.
Long answer: use metaprogramming!
Because Javascript does not have Unicode regexes, I wrote a Python program to iterate over the whole of Unicode and filter by Unicode name. It is difficult to get this right manually. Why not let the computer do the dirty and menial work?
import unicodedata
import re
import sys
def unicodeNameMatch(pattern, codepoint):
try:
return re.match(pattern, unicodedata.name(unichr(codepoint)), re.I)
except ValueError:
return None
def regexChr(codepoint):
return chr(codepoint) if 32 <= codepoint < 127 else "\\u%04x" % codepoint
names = sys.argv
prev = None
js_regex = ""
for codepoint in range(pow(2, 16)):
if any([unicodeNameMatch(name, codepoint) for name in names]):
if prev is None: js_regex += regexChr(codepoint)
prev = codepoint
else:
if not prev is None: js_regex += "-" + regexChr(prev)
prev = None
print "[" + js_regex + "]"
Invoke it like this: python char_class.py latin digit and you get the character class mentioned above. It's an ugly char class but you know for sure that you catched all characters whose names contain latin or digit.
Browse the Unicode Character Database to view the names of all unicode characters. The name is in uppercase after the first semicolon, for example for A its the line
0041;LATIN CAPITAL LETTER A;Lu;0;L;;;;;N;;;;0061;
Try python char_class.py "latin small" and you get a character class for all latin small letters.
Edit: There is a small misfeature (aka bug) in that \u271d-\u271d occurs in the regex. Perhaps this fix helps: Replace
if not prev is None: js_regex += "-" + regexChr(prev)
by
if not prev is None and prev != codepoint: js_regex += "-" + regexChr(prev)

var desired = stringToReplace.replace(/[\u0000-\u007F][\W]/gi, '');
might do the trick.
See also this Javascript + Unicode regexes question.

If you must insist on whitelisting here is the rawest way of doing it:
Test if string contains only letters (a-z + é ü ö ê å ø etc..)
It works by keeping track of 'all' unicode letter chars.

Unfortunately, Javascript does not support Unicode character properties (which would be just the right regex feature for you). If changing the language is an option for you, PHP (for example) can do this:
preg_replace("/[^\pL0-9_\s]/", "", $str);
Where \pL matches any Unicode character that represents a letter (lower case, upper case, modified or unmodified).
If you have to stick with JavaScript and cannot use the library suggested by Tim Down, the only options are probably either blacklisting or whitelisting. But your bounty mentions that blacklisting is not actually an option in your case. So you will probably simply have to include the special characters from your relevant language manually. So you could simply do this:
var desired = stringToReplace.replace(/[^\w\sñáóí]/gi, '');
Or use their corresponding Unicode sequences:
var desired = stringToReplace.replace(/[^\w\s\u00F1\u00C1\u00F3\u00ED]/gi, '');
Then simply add all the ones you want to take care of. Note that the case-insensitive modifier also works with Unicode sequences.

Javascript: String replace problem

I've got a string which contains q="AWORD" and I want to replace q="AWORD" with q="THEWORD". However, I don't know what AWORD is.. is it possible to combine a string and a regex to allow me to replace the parameter without knowing it's value? This is what I've got thus far...
globalparam.replace('q="/+./"', 'q="AWORD"');

What you have is just a string, not a regular expression. I think this is what you want:
globalparam.replace(/q=".+?"/, 'q="THEWORD"');
I don't know how you got the idea why you have to "combine" a string and a regular expression, but a regex does not need to exist of wildcards only. A regex is like a pattern that can contain wildcards but otherwise will try to match the exact characters given.
The expression shown above works as follows:
q=": Match the characters q, = and ".
.+?": Match any character (.) up to (and including) the next ". There must be at least one character (+) and the match is non-greedy (?), meaning it tries to match as few characters as possible. Otherwise, if you used .+", it would match all characters up to the last quotation mark in the string.
Learn more about regular expressions.

Felix's answer will give you the solution, but if you actually want to construct a regular expression using a string you can do it this way:
var fullstring = 'q="AWORD"';
var sampleStrToFind = 'AWORD';
var mat = 'q="'+sampleStrToFind+'"';
var re = new RegExp(mat);
var newstr = fullstring.replace(re,'q="THEWORD"');
alert(newstr);
mat = the regex you are building, combining strings or whatever is needed.
re = RegExp constructor, if you wanted to do global, case sensitivity, etc do it here.
The last line is string.replace(RegExp,replacement);

Javascript string replace with regex to strip off illegal characters

Need a function to strip off a set of illegal character in javascript: |&;$%#"<>()+,
This is a classic problem to be solved with regexes, which means now I have 2 problems.
This is what I've got so far:
var cleanString = dirtyString.replace(/\|&;\$%#"<>\(\)\+,/g, "");
I am escaping the regex special chars with a backslash but I am having a hard time trying to understand what's going on.
If I try with single literals in isolation most of them seem to work, but once I put them together in the same regex depending on the order the replace is broken.
i.e. this won't work --> dirtyString.replace(/\|<>/g, ""):
Help appreciated!

What you need are character classes. In that, you've only to worry about the ], \ and - characters (and ^ if you're placing it straight after the beginning of the character class "[" ).
Syntax: [characters] where characters is a list with characters.
Example:
var cleanString = dirtyString.replace(/[|&;$%#"<>()+,]/g, "");

I tend to look at it from the inverse perspective which may be what you intended:
What characters do I want to allow?
This is because there could be lots of characters that make in into a string somehow that blow stuff up that you wouldn't expect.
For example this one only allows for letters and numbers removing groups of invalid characters replacing them with a hypen:
"This¢£«±Ÿ÷could&*()\/<>be!##$%^bad".replace(/([^a-z0-9]+)/gi, '-');
//Result: "This-could-be-bad"

You need to wrap them all in a character class. The current version means replace this sequence of characters with an empty string. When wrapped in square brackets it means replace any of these characters with an empty string.
var cleanString = dirtyString.replace(/[\|&;\$%#"<>\(\)\+,]/g, "");

Put them in brackets []:
var cleanString = dirtyString.replace(/[\|&;\$%#"<>\(\)\+,]/g, "");