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Replacing only exact matches in Regex [duplicate]

This question already has answers here:
How can I match a whole word in JavaScript?
(4 answers)
Closed 4 months ago.
I have a javascript function that replaces strings that match against a profanity library with symbols.
eg. damn is transformed into ##$%&!
However, this replace is non-discriminating, and will transform partial matches as well.
eg. classic is transformed into cl##$%&!ic
I'm very new to TS/JS and Regex and would like some help understanding this function's arguments and how to modify them for only exact matches.
this.replace(new RegExp(str1.replace(/([\/\,\!\\\^\$\{\}\[\]\(\)\.\*\+\?\|\<\>\-\&])/g, "\\$&"), (ignore ? "gi" : "g")), (typeof (str2) == "string") ? str2.replace(/\$/g, "$$$$") : str2);

To avoid partial matches, the normal solution is to surround what you want to match with word boundaries \b.
The following example assumes that profanities does not contain any words that contain regex special characters.
Notice that the "shit" in "mishit" and the "ass" in "class" do not get replaced.
const profanities = ['damn', 'shit', 'ass'];
// As regex literal: /\b(?:damn|shit|ass)\b/
const re = new RegExp('\\b(?:' + profanities.join('|') + ')\\b', 'gi');
const text = "Damn, another mishit. I am shit at pool. Time to get my ass to class.";
console.log(text.replace(re, '##$%&!'));

How to check pipeline and comma (|,) in a string using Regex [duplicate]

This question already has an answer here:
All characters in a string must match regex
(1 answer)
Closed 3 years ago.
I have to check whether a given string contains pipeline and comma (|,) in a string and throw error if it contain the same.throwing error part i can handle but facing issue while creating regex for this.
i have used this regex to validate it [^|,] but its not working.
String is "Equinix | Megaport" but my vregex is not throwing error as it contains | in it

In JavaScript, try:
var subject = "Equinix | Megaport";
if (/^[^|,]+$/.test(subject)) {
document.write('No bad character(s)');
} else {
document.write('Contains bad character(s)');
};
Also note that I slightly changed your pattern to include the + operator after your negated character class, which means match the character class between one and unlimited times.

simple regular expression with equals but dont want to use split [duplicate]

This question already has answers here:
Regular Expressions in JavaScript for URL Capture
(4 answers)
Closed 7 years ago.
I have the message string as follows.
string=052
I need to use regular expression not split.
I want to return everything past the equals. 052
This is what i tried and gives me id=null
var regex = '[^?string=]';
var id2 = mystring.match(regex);
I have tried online regex checkers and it looks like it matches all but the a
is there a better reg ex i should try? id should not equal null.

You're using String.match() incorrectly. Try this:
var regex = '^message=(.*)$';
var id = queryString.match(regex)[1];
.match() returns an array; the first element (at [0]) is the entire matched string, and the second element (at [1]) is the part that's matched in the (first) set of parentheses in the regex.

RegEx is not working with backslash in javascript [duplicate]

This question already has answers here:
Why does a RegExp with global flag give wrong results?
(7 answers)
Closed 7 years ago.
I am trying to validate strings with following pattern.
starts with 3 digits followed by hyphen followed by 2 digits and ends with 4 digits.
eg : 123-45-6789
I need to pass the pattern to a RegExp object as a string variable (I am getting the pattern value from a html element attribute ), hence I have escaped the RegExp with double slashes. To my surprise, I am seeing weird results.
var pattern = "^\\d{3}-\\d{2}-\\d{4}$";
var inputRegex = new RegExp(pattern, "g");
console.log(inputRegex.test('132-45-7899')); //True
console.log(inputRegex.test('132-45-7899')); //False
console.log(inputRegex.test('132-45-7899')); //True
console.log(inputRegex.test('132-45-7899')); //False
What am I doing wrong here? and what is the reason for this inconsistent results?
EDIT:
Now I understood why the results are inconsistent , but my actual problem remained unsolved.
HTML elem is like below
SCENARIO 1:
<input data-val-regexp="^\\d{3}-\\d{2}-\\d{4}$" id="ticketNumber">
var regexPattern = input.attr("data-val-regexp");
var inputRegex = new RegExp(regexPattern);
return inputRegex.test('123-45-6789'); // False
When I inspected, the inputRegex.source is "^\\d{3}-\\d{2}-\\d{4}$"
SCENARIO 2:
If I assign the regexPattern with string literals everything works fine.
var pattern = "^\\d{3}-\\d{2}-\\d{4}$";
var inputRegex = new RegExp(pattern);
console.log(inputRegex.test('132-45-7899')); //True
In the above case inputRegex source value is "^\d{3}-\d{2}-\d{4}$"
Why is the backslash escaped in the second scenario, where as it didn't in the first scenario? What needs to be done in order to make scenario 1 work?

Because you're creating the regular expression with the g flag, it remembers where it left off and starts matching there, instead of at the beginning, the next time you call test. See Why does Javascript's regex.exec() not always return the same value?.

passing variable to a regexp in javascript [duplicate]

This question already has answers here:
How do you use a variable in a regular expression?
(27 answers)
Is there a RegExp.escape function in JavaScript?
(18 answers)
Closed 4 years ago.
Possible Duplicate:
Escape string for use in Javascript regex
I have a msg like this:
Max {0} chars allowed in {1}
And I have a function to create a message using the arguments passed as
for(var i = 0; i < agrs.length; i++){
reg = new RegExp('\{'+i+'\}', 'gi');
key = key.replace(reg,agrs[i])
}
The problem is that it's not able to take the param i to create the reg exp.
What's the way to achieve this?

Your regexp is /{0}/gi since you create it from a string. And it is not a valid expression.
You need to escape { in the regexp because it has a special meaning in the regexp syntax, so it should be:
new RegExp('\\{'+i+'\\}', 'gi');
which is /\\{0\\}/gi. You need to escape the escaping \\ in the string.

I would strongly encourage you to use the functional form of String.replace() to solve your problem, rather than trying to parametrize the regexp in a for-loop that iterates over {0},{1},etc.
In other words, rather than look for '{0}' or '{1}', just look for '{([0-9]+)}' (i.e. {}'s surrounding an arbitrary number, and pass a function to the replace() function to intelligently replace these expressions based on the number between the {}'s. This lets you use a RegExp literal which you can write as /{([0-9]+)}/ rather than mess around with escape characters in strings.
Something like this:
s='the song says {0} little {1} little {2} little-endians';
args=['zero','one','two'];
new_string = s.replace(/{([0-9]+)}/g, function(wholematch,firstmatch)
{return args[(+firstmatch)]; }
);
which yields
the song says zero little one little two little-endians
See this similar question.
edit: if you want to leave alone items that are greater than the length of your args list, make sure to sanity-check the parameter number:
s='the song says {0} little {1} little {2} little-endians,\n'+
' {3} little {4} little {5} little-endians';
args=['zero','one','two'];
new_string = s.replace(/{([0-9]+)}/g, function(wholematch,firstmatch)
{var i = +firstmatch; return i < args.length ? args[i] : wholematch;}
);
which yields
the song says zero little one little two little-endians,
{3} little {4} little {5} little-endians

I think your second line is supposed to be
reg = new RegExp('\{'+i+'\}', 'gi');
but I don't know if that's your only problem or not.

function escapeRegExp(str) {
return str.replace(/[-\[\]\/\{\}\(\)\*\+\?\.\\\^\$\|]/g, "\\$&");
}
var re = new RegExp(escapeRegExp(str));
See: Escape string for use in Javascript regex