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Odd RegEx request for Javascript

I'm having trouble with a certain RegEx replacement string for later use in Javascript.
We have quite a bit of text that was stored in a rather odd format that we aren't allowed to fix.
But we do need to find all the "network path" strings inside it, following these rules:
A. The matches always start with 2 backslashes.
B. The matching characters should stop as soon as it hits a first occurrence of any 1 of these:
A < character
A space
A line feed
A carriage return
A & character
A literal "\r" or "\n" string (but only if occurring at end of line)
We "almost" have it working with /\\\\[^ &<\s]*/gi as shown in this RegEx Tester page:
https://regex101.com/r/T4cDOL/5
Even if we get it working, the RegEx has to be even futher "escape escaped" before putting on
our Javascript code, but that's also not working as expected.

From your example, it seems you literally have a backslash followed by an n and a backslash followed by an r (as opposed to a newline or carriage return), which means you can't only use a negated character class (since you need to handle a sequence of two characters). I'd use a positive lookahead to know where to stop, so I can use an alternation for that part.
You haven't said what parts of those strings should match, so I've had to guess a bit, but here's my best guess (with useful input from Niet the Dark Absol):
const rex = /\\\\.*?(?=[ &<\r\n]|\\[rn](?:$| ))/gmi;
That says:
Match starting with \\
Take everything prior to the lookahead (non-greedy)
Lookahead: An alternation of:
A space, &, <, carriage return (\r, character 13), or a newline (\n, character 10); or
A backslash followed by r or n if that's either at the end of a line or followed by a space (so we get the \nancy but not the \n after it).
Updated regex101
You might want to have more characters than just a space after the \r/\n. If so, make it a character class (and/or use \s for "whitespace" if that applies):
const rex = /\\\\.*?(?=[ &<\r\n]|\\[rn](?:$|[ others]))/gmi;
// −−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−^^^^^^^^^

what is max length of string to test regular expression in javascript

I am using following regex if I pass lenghty string to check, regex101.com showing timeout message. Is there any ideal length to test regular expresssion?
^(\d+[\s\r\n\t,]*){1,100}$
https://regex101.com/r/eC5qO7/1

I suggest running a split, then making sure what is split is a number, so:
var test = "123,456 789 101112 asdf";
var numbers = test.split(/\s*,\s*|\s+/);
numbers.forEach(function(n) {
if (n % 1 !== 0) {
alert(n + " is not an integer");
}
});

The catastrophic backtracking is caused by the fact that the [\s\r\n\t,] character class has a * quantifier applied to it, and then a + quantifier is set to the whole group. The regex backtracks into each digit to search for optional whitespace and commas, which creates a huge number of possibilities that the engine tries before running into the "catastrophe".
Besides, there is another potential bottleneck: \s in the character class can also match \r, \n and \t. See see Do character classes present the same risk?.
Without atomic groups and possessive quantifiers, the regex optimization is only possible by means of making one of the "separators" obligatory. In this case, it is clearly a comma (judging by the example string). Since you just want to validate the number of input numbers separated with commas and optional spaces, you can use a simpler regex:
^(?:[0-9]+\s*,\s*){1,100}$
Here, it fails gracefully, and here it matches the string OK.
If a comma at the end is optional, use
^(?:\d+\s*,\s*){1,99}\d+,?\s*$
See demo
Also note you do not need the i modifier, as there are no letters in the pattern.

Regex to allow certain special characters - escape issue

I've working on a javascript regex that I intend to use with the jquery validate plugin (I'll add this as an additional method). It must (among other rules):
test if at least one of the following special characters is entered:
!, ", #, $, %, &, ', (, ), *, +,-, .,/, :, ;, <, =, >, ?, #, [, \, ], ^, _, `, {, |, }, ~
not allow 3 consecutive identical characters:
passed:
aa
99
++
not passed:
aaa
999
+++
The problem with my regex is that is having problem with these mentioned rules:
I think the issue is related to escaping and I've tried escaping + and - to no avail. Can anyone help! This is my regex: http://regexr.com/3ack3

This is one of those requirements where you can really simplify your life by using multiple regexes, rather than trying to cram all the logic into one complex regex with many assertions. Here's some JavaScript that implements your requirement:
var specialCharRegex = /[!"#$%&'()*+.\/:;<=>?#\[\\\]^_`{|}~-]/;
var threeConsecutiveRegex = /(.)\1\1/;
var input = prompt();
if (specialCharRegex.test(input) && !threeConsecutiveRegex.test(input)) {
alert('passed');
} else {
alert('failed');
} // end if
http://jsfiddle.net/t8609xv2/
Some notes on the trickier points:
inside the bracket expression, the following four special characters had to be backslash-escaped: /[\]. (Forward slash because it delimits the regex, backslash because it's the escape character, and the brackets because they delimit the bracket expression.)
inside the bracket expression, the dash had to be moved to the end, because otherwise it would likely specify a character range. When at the end, it never specifies a range, so it's always safer to put it there.
This modular approach also benefits maintainability, as you will more easily be able to make changes (modify/add/remove regexes, or change the if-test logic) at a later point in time.
Another benefit is that you could test each regex independently, which could allow you to provide a more accurate error message to the user, as opposed to just saying something like "invalid password".
Edit: Here's how you can whitelist the chars that are accepted in the input:
var specialCharRegex = /[!"#$%&'()*+.\/:;<=>?#\[\\\]^_`{|}~-]/;
var threeConsecutiveRegex = /(.)\1\1/;
var nonWhitelistCharRegex = /[^a-zA-Z0-9!"#$%&'()*+.\/:;<=>?#\[\\\]^_`{|}~-]/;
var input = prompt();
if (specialCharRegex.test(input) && !threeConsecutiveRegex.test(input) && !nonWhitelistCharRegex.test(input)) {
alert('passed');
} else {
alert('failed');
} // end if
http://jsfiddle.net/t8609xv2/2/

^(?=.*[!"#$%&'()*+,,\/:;<=>?#\[\]^_`{|}~-])(?!.*(.)\1\1).*$
Try this.See demo.
https://regex101.com/r/wX9fR1/10
You need a positive lookahead to check for special characters.
And
A negative lookahead to check if a character is is there 3 times.

You can use this regex:
^(?!.*?(.)\1{2})(?=[^a-z]*[a-z])(?=[^A-Z]*[A-Z])(?=\D*\d)(?=.*?[!##$%^&*()_=\[\]{};':"\\|,.<>\/?+-]).{8,20}$
RegEx Demo
You might be able to shorten it using:
^(?!.*?(.)\1{2})(?=[^a-z]*[a-z])(?=[^A-Z]*[A-Z])(?=\D*\d)(?=.*?[\W_]).{8,20}$
i.e. using non-word property \W instead of listing each and every special character.

Regex-Groups in Javascript

I have a problem using a Javascript-Regexp.
This is a very simplified regexp, which demonstrates my Problem:
(?:\s(\+\d\w*))|(\w+)
This regex should only match strings, that doesn't contain forbidden characters (everything that is no word-character).
The only exception is the Symbol +
A match is allowed to start with this symbol, if [0-9] is trailing.
And a + must not appear within words (44+44 is not a valid match, but +4ad is)
In order to allow the + only at the beginning, I said that there must be a whitespace preceding. However, I don't want the whitespace to be part of the match.
I tested my regex with this tool: http://regex101.com/#javascript and the resultig matches look fine.
There are 2 Issues with that regexp:
If I use it in my JS-Code, the space is always part of the match
If +42 appears at the beginning of a line, it won't be matched
My Questions:
How should the regex look like?
Why does this regex add the space to the matches?
Here's my JS-Code:
var input = "+5ad6 +5ad6 sd asd+as +we";
var regexp = /(?:\s(\+\d\w*))|(\w+)/g;
var tokens = input.match(regexp);
console.log(tokens);

How should the regex look like?
You've got multiple choices to reach your goal:
It's fine as you have it. You might allow the string beginning in place of the whitespace as well, though. Just get the capturing groups (tokens[1], tokens[2]) out of it, which will not include the whitespace.
If you didn't use JavaScript, a lookbehind could help. Unfortunately it's not supported.
Require a non-word-boundary before the +, which would make every \w character before the + prevent the match:
/\B\+\d\w+|\w+/
Why does this regex add the space to the matches?
Because the regex does match the whitespace. It does not add the \s(\+\d\w+) to the captured groups, though.

I want to ignore square brackets when using javascript regex [duplicate]

This question already has answers here:
Why is this regex allowing a caret?
(3 answers)
Closed 1 year ago.
I am using javascript regex to do some data validation and specify the characters that i want to accept (I want to accept any alphanumeric characters, spaces and the following !&,'\- and maybe a few more that I'll add later if needed). My code is:
var value = userInput;
var pattern = /[^A-z0-9 "!&,'\-]/;
if(patt.test(value) == true) then do something
It works fine and excludes the letters that I don't want the user to enter except the square bracket and the caret symbols. From all the javascript regex tutorials that i have read they are special characters - the brackets meaning any character between them and the caret in this instance meaning any character not in between the square brackets. I have searched here and on google for an explanation as to why these characters are also accepted but can't find an explanation.
So can anyone help, why does my input accept the square brackets and the caret?

The reason is that you are using A-z rather than A-Za-z. The ascii range between Z (0x5a) and a (0x61) includes the square brackets, the caret, backquote, and underscore.

Your regex is not in line with what you said:
I want to accept any alphanumeric characters, spaces and the following !&,'\- and maybe a few more that I'll add later if needed
If you want to accept only those characters, you need to remove the caret:
var pattern = /^[A-Za-z0-9 "!&,'\\-]+$/;
Notes:
A-z also includesthe characters: [\]^_`.
Use A-Za-z or use the i modifier to match only alphabets:
var pattern = /^[a-z0-9 "!&,'\\-]+$/i;
\- is only the character -, because the backslash will act as special character for escaping. Use \\ to allow a backslash.
^ and $ are anchors, used to match the beginning and end of the string. This ensures that the whole string is matched against the regex.
+ is used after the character class to match more than one character.
If you mean that you want to match characters other than the ones you accept and are using this to prevent the user from entering 'forbidden' characters, then the first note above describes your issue. Use A-Za-z instead of A-z (the second note is also relevant).

I'm not sure what you want but I don't think your current regexp does what you think it does:
It tries to find one character is not A-z0-9 "!&,'\- (^ means not).
Also, I'm not even sure what A-z matches. It's either a-z or A-Z.
So your current regexp matches strings like "." and "Hi." but not "Hi"

Try this: var pattern = /[^\w"!&,'\\-]/;
Note: \w also includes _, so if you want to avoid that then try
var pattern = /[^a-z0-9"!&,'\\-]/i;
I think the issue with your regex is that A-z is being understood as all characters between 0x41 (65) and 0x7A (122), which included the characters []^_` that are between A-Z and a-z. (Z is 0x5A (90) and a is 0x61 (97), which means the preceding characters take up 0x5B thru 0x60).