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RegEx for replacing punctuation excluding negative numbers

Currently, to remove punctuation from a string, I use:
export function scrubPunctuation(text) {
let reg = /\b[-.,()&$#![\]{}"']+\B|\B[-.,()&$#![\]{}"']+\b/g;
return text.replace(reg, "");
}
but this also removes -1, where - is not so much "punctuation" as part of a numerical value.
How do I solve this problem?
Example use case:
I have take a string from a user that might look like this:
const userStr = " I want something, sort of, that has at least one property < -1.02 ? "
Currently, my approach is to first trim the string to remove the leading / trailing white space.
Then I "scrub" punctuation from the string.
From the example of userStr above, I might eventually parse out (via some unrelated to regex):
const relevant = ["something", "at least one", "<", "-1.02"]
In general, non-numeric punctuation is irrelevant.

Split your first character set. Remove the hyphen from the first set and add a Negative lookahead for the hyphen:
[-]+(?![0-9]) \\a Hyphen not followed by a number
And the full expression:
\b[-]+(?![0-9])|[-.,()&$#![\]{}"']+\B|\B[.,()&$#![\]{}"']+\b
Here is a working example

If you don't want the minus sign or the dot or comma removed form the digits, one option might be to capture what you want to keep (in this case a digit with an optional decimal part) and match what you want to remove.
(-?\d+(?:[.,]\d+)*)|[-.,()&$#![\]{}"']+
Regex demo
let pattern = /(-?\d+(?:[.,]\d+)*)|[-.,()&$#![\]{}"']+/g;
let str = "This is -4, -55 or -4,00.00 (test) 5,00";
let res = str.replace(pattern, "$1");
console.log(res);

something like /[,?!.']/g could do the job and you add whatever you want
const text = "bar........,foo,????!-1'poo!!!?'";
const res = text.replace(/[,?!.']/g, "")
console.log(res)

I would split it into two.
First I would remove everything but alphanumeric and -.
/[^a-z0-9\-\s\n]/gi
It is a little more readable than your method and should give the same result unless there is some character you want to keep (like whitespace \s and newline \n).
To get rid of the punctuation "-", I would use:
/-(\d*)/g
So altogether:
export function scrubPunctuation(text) {
let reg = /[^a-z0-9\-\s\n]/gi;
let reg2 = /-(\d*)/g;
text = text.replace(reg, "");
return text.replace(reg2, "$1");
}
Haven't tested it, but it should work

String.prototype.replace() to remove dashes and underscores

I'm trying to remove all ocurrences of dashes and underscores of a string with String.prototype.replace(), but it's not working and I don't know why. My code:
var str = "dash-and_underscore";
str = str.replace(/_|\-/, " ");
console.log(str);
outputs:
"dash and_underscore"
in the Chrome console.
Since the | acts like the OR opperator, what am I doing wrong? I've tried the solution here, but it didn't work, or I'm too dumb to understand - which is an option ;)

Try this:
str = str.replace(/[_-]/g, " ");
[..] defines a character class
g means global research
(You can write it with a quantifier /[_-]+/g to remove several consecutive characters at a time.)
or
str = str.replace(/_|-/g, " ");
that is correct too, but slower. Note that the dash doesn't need to be escaped out of a character class since it isn't a special character.

JavaScript: How can I remove any words containing (or directly preceding) capital letters, numbers, or commas, from a string?

I'm trying to write the code so it removes the "bad" words from the string (the text).
The word is "bad" if it has comma or any special sign thereafter. The word is not "bad" if it contains only a to z (small letters).
So, the result I'm trying to achieve is:
<script>
String.prototype.azwords = function() {
return this.replace(/[^a-z]+/g, "0");
}
var res = "good Remove remove1 remove, ### rem0ve? RemoVE gooood remove.".azwords();//should be "good gooood"
//Remove has a capital letter
//remove1 has 1
//remove, has comma
//### has three #
//rem0ve? has 0 and ?
//RemoVE has R and V and E
//remove. has .
alert(res);//should alert "good gooood"
</script>

Try this:
return this.replace(/(^|\s+)[a-z]*[^a-z\s]\S*(?!\S)/g, "");
It tries to match a word (that is surrounded by whitespaces / string ends) and contains any (non-whitespace) character but at least one that is not a-z. However, this is quite complicated and unmaintainable. Maybe you should try a more functional approach:
return this.split(/\s+/).filter(function(word) {
return word && !/[^a-z]/.test(word);
}).join(" ");

okay, first off you probably want to use the word boundary escape \b in your regex. Also, it's a bit tricky if you match the bad words, because a bad word might contain lower case chars, so your current regex will exclude anything which does have lowecase letters.
I'd be tempted to pick out the good words and put them in a new string. It's a much easier regex.
/\b[a-z]+\b/g
NB: I'm not totally sure that it'll work for the first and last words in the string so you might need to account for that as well. http://www.regextester.com/ is exceptionally useful.
EDIT: as you want punctiation after the word to be 'bad', this will actually do what I was suggesting
(^|\s)[a-z]+(\s|$)

Firstly I wouldn't recommend changing the prototype of String (or of any native object) if you can avoid because you leave yourself open to conflicts with other code that might define the same property in different ways. Much better to put custom methods like this on a namespaced object, though I'm sure some will disagree.
Second, is there any need to use RegEx completely? (Genuine question; not trying to be facetious.)
Here is an example of the function with plain old JS using a little bit of RegEx here and there. Easier to comment, debug, and reuse.
Here is the code:
var azwords = function(str) {
var arr = str.split(/\s+/),
len = arr.length,
i = 0,
res = "";
for (i; i < len; i += 1) {
if (!(arr[i].match(/[^a-z]/))) {
res += (!res) ? arr[i] : " " + arr[i];
}
}
return res;
}
var res = "good Remove remove1 remove, ### rem0ve? RemoVE gooood remove."; //should be "good gooood"
//Remove has a capital letter
//remove1 has 1
//remove, has comma
//### has three #
//rem0ve? has 0 and ?
//RemoVE has R and V and E
//remove. has .
alert(azwords(res));//should alert "good gooood";

Try this one:
var res = "good Remove remove1 remove, ### rem0ve? RemoVE gooood remove.";
var new_one = res.replace(/\s*\w*[#A-Z0-9,.?\\xA1-\\xFF]\w*/g,'');
//Output `good gooood`
Description:
\s* # zero-or-more spaces
\w* # zero-or-more alphanumeric characters
[#A-Z0-9,.?\\xA1-\\xFF] # matches any list of characters
\w* # zero-or-more alphanumeric characters
/g - global (run over all string)

This will find all the words you want /^[a-z]+\s|\s[a-z]+$|\s[a-z]+\s/g so you could use match.
this.match(/^[a-z]+\s|\s[a-z]+$|\s[a-z]+\s/g).join(" "); should return the list of valid words.
Note that this took some time as a JSFiddle so it maybe more efficient to split and iterate your list.

Enhancing regex of thousands separator?

I saw this beautiful script to add thousands separator to js numbers:
function thousandSeparator(n, sep)
{
var sRegExp = new RegExp('(-?[0-9]+)([0-9]{3})'),
sValue = n + '';
if(sep === undefined)
{
sep = ',';
}
while(sRegExp.test(sValue))
{
sValue = sValue.replace(sRegExp, '$1' + sep + '$2');
}
return sValue;
}
usage :
thousandSeparator(5000000.125, '\,') //"5,000,000.125"
However I'm having a trouble accepting the while loop.
I was thinking to change the regex to : '(-?[0-9]+)([0-9]{3})*' asterisk...
but now , how can I apply the replace statement ?
now I will have $1 and $2..$n
how can I enhance the replace func?
p.s. the code is taken from here http://www.grumelo.com/2009/04/06/thousand-separator-in-javascript/

There is no need to use replace, you can just add toLocaleString instead:
console.log((5000000.125).toLocaleString('en'));
More information: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Number/toLocaleString

Your assumption
now i will have $1 and $2..$n
is wrong. You have two groups, because you have two sets of brackets.
(-?[0-9]+)([0-9]{3})*
1. ^^^^^^^^^^
2. ^^^^^^^^^^
And then you repeat the second group. If it matches the second time, it overwrites the result of the first match, when it matches the third time, it overwrites ...
That means when matching is complete, $2 contains the value of the last match of that group.
First approach
(\d)(?=(?:[0-9]{3})+\b)
and replace with
$1,
See it on Regexr
It has the flaw that it does insert the comma also on the right of the dot. (I am working on it.)
Second approach
(\d)(?:(?=\d+(?=[^\d.]))(?=(?:[0-9]{3})+\b)|(?=\d+(?=\.))(?=(?:[0-9]{3})+(?=\.)))
and replace with
$1,
See it on Regexr
So now its getting a bit more complicated.
(\d) # Match a digit (will be reinserted)
(?:
(?=\d+(?=[^\d.])) # Use this alternative if there is no fractional part in the digit
(?=(?:\d{3})+ # Check that there are always multiples of 3 digits ahead
\b) # Till a word boundary
| # OR
(?=\d+(?=\.)) # There is a fractional part
(?=(?:\d{3})+ # Check that there are always multiples of 3 digits ahead
(?=\.)) # Till a dot
)
Problem:
does also match the fractional part if there is not the end of the string following.

Here is an ugly script to contrast your beautiful script.
10000000.0001 .toString().split('').reverse().join('')
.replace(/(\d{3}(?!.*\.|$))/g, '$1,').split('').reverse().join('')
Since we don't have lookbehinds, we can cheat by reversing the string and using lookaheads instead.
Here it is again in a more palatable form.
function thousandSeparator(n, sep) {
function reverse(text) {
return text.split('').reverse().join('');
}
var rx = /(\d{3}(?!.*\.|$))/g;
if (!sep) {
sep = ',';
}
return reverse(reverse(n.toString()).replace(rx, '$1' + sep));
}

How about this one:
result = "1235423.125".replace(/\B(?=(\d{3})+(?!\d))/g, ',') //1,235,423.125

Try this one:
result = subject.replace(/([0-9]+?)([0-9]{3})(?=.*?\.|$)/mg, "$1,$2");
Test here

Replace multiple whitespaces with single whitespace in JavaScript string

I have strings with extra whitespace characters. Each time there's more than one whitespace, I'd like it be only one. How can I do this using JavaScript?

Something like this:
var s = " a b c ";
console.log(
s.replace(/\s+/g, ' ')
)

You can augment String to implement these behaviors as methods, as in:
String.prototype.killWhiteSpace = function() {
return this.replace(/\s/g, '');
};
String.prototype.reduceWhiteSpace = function() {
return this.replace(/\s+/g, ' ');
};
This now enables you to use the following elegant forms to produce the strings you want:
"Get rid of my whitespaces.".killWhiteSpace();
"Get rid of my extra whitespaces".reduceWhiteSpace();

Here's a non-regex solution (just for fun):
var s = ' a b word word. word, wordword word ';
// with ES5:
s = s.split(' ').filter(function(n){ return n != '' }).join(' ');
console.log(s); // "a b word word. word, wordword word"
// or ES2015:
s = s.split(' ').filter(n => n).join(' ');
console.log(s); // "a b word word. word, wordword word"
Can even substitute filter(n => n) with .filter(String)
It splits the string by whitespaces, remove them all empty array items from the array (the ones which were more than a single space), and joins all the words again into a string, with a single whitespace in between them.

using a regular expression with the replace function does the trick:
string.replace(/\s/g, "")

I presume you're looking to strip spaces from the beginning and/or end of the string (rather than removing all spaces?
If that's the case, you'll need a regex like this:
mystring = mystring.replace(/(^\s+|\s+$)/g,' ');
This will remove all spaces from the beginning or end of the string. If you only want to trim spaces from the end, then the regex would look like this instead:
mystring = mystring.replace(/\s+$/g,' ');
Hope that helps.

jQuery.trim() works well.
http://api.jquery.com/jQuery.trim/

I know I should not necromancy on a subject, but given the details of the question, I usually expand it to mean:
I want to replace multiple occurences of whitespace inside the string with a single space
...and... I do not want whitespaces in the beginnin or end of the string (trim)
For this, I use code like this (the parenthesis on the first regexp are there just in order to make the code a bit more readable ... regexps can be a pain unless you are familiar with them):
s = s.replace(/^(\s*)|(\s*)$/g, '').replace(/\s+/g, ' ');
The reason this works is that the methods on String-object return a string object on which you can invoke another method (just like jQuery & some other libraries). Much more compact way to code if you want to execute multiple methods on a single object in succession.

var x = " Test Test Test ".split(" ").join("");
alert(x);

Try this.
var string = " string 1";
string = string.trim().replace(/\s+/g, ' ');
the result will be
string 1
What happened here is that it will trim the outside spaces first using trim() then trim the inside spaces using .replace(/\s+/g, ' ').

How about this one?
"my test string \t\t with crazy stuff is cool ".replace(/\s{2,9999}|\t/g, ' ')
outputs "my test string with crazy stuff is cool "
This one gets rid of any tabs as well

If you want to restrict user to give blank space in the name just create a if statement and give the condition. like I did:
$j('#fragment_key').bind({
keypress: function(e){
var key = e.keyCode;
var character = String.fromCharCode(key);
if(character.match( /[' ']/)) {
alert("Blank space is not allowed in the Name");
return false;
}
}
});
create a JQuery function .
this is key press event.
Initialize a variable.
Give condition to match the character
show a alert message for your matched condition.