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How to show first 2 characters and replace all the last by * [closed]

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I'm wondering how to show the first two characters and replace all last character of a string by symbol *.
Ex: 121,121,121 -> 12x,xxx,xxx .
Thanks

I love using regex when it comes to replace string according to some pattern.
var p = '121,121,121';
var regex = /(?<=.{2})([0-9])/gm;
console.log(p.replace(regex, 'x'));

You can use substring and regular expression. See the sample below.
var str = "121,121,121";
var res = str.substring(0, 2) + '' + str.substring(2, str.length).replace(/[0-9]/g,"x");
alert(res);

Just use substring and replace with a simple regex (to single out digits and keep commas and other punctuation):
const str = "121,121,121";
const obfuscated = `${str.substring(0, 2)}${str.substring(2).replace(/\d/g, "*")}`;
console.log(obfuscated);

Replace an expression in string using javascript [closed]

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I have a string that has some numbers in the middle of the string.
For example,
var str = "abcd-123456.com"
I want to remove the numbers like this
abcd.com
I am not trying to replace all numbers.
I have to replace only -*. expression with "".
How do I do this in JavaScript?

var str = "abcd-123456.com"
str = str.replace(/-[0-9]*/g, '')

Your comments lead me to believe you want
var str = "abcd-123456.com";
var str1 = str.substring(0,str.indexOf("-"))+str.substring(str.indexOf("."))
console.log(str1);
//or with regex
// dasah plus 6 digits to nothing
var str2 = str.replace(/-\d{6}/,"")
console.log(str2);
// dash, digits and an dot to dot
var str3 = str.replace(/-\d+\./,".")
console.log(str3);

AS PER YOUR COMMENT
The answer of Shivaji Varma will nearly do the tricks.
var str = "abc5d-123456.c0om"
str = str.replace(/-[0-9]*./g, "")
console.log(str)
Soooo,
Using the slash indicate a regular expression.
[0-9] indicate you want to replace all digit between 0 and 9 (included)
"*" to remove all digits
"-" and "." to delimite

Need to test JS regex [closed]

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When I type in my input the text appears on image. I want to make the code work like that:
When user types character that is recognized by regex then place it on image else just throw some error or do not let it type at all in the input field.
This is the regex: [A-Z0-9a-z&,.-/()#*+!?"':; -]
I'm trying to achieve it like this:
$("#firstText").keyup(function () {
var value = $(this).val().toUpperCase();
var regex = "[A-Z0-9a-z&,.-/()#*+!?"':; -]";
if(regex.test(value))
{
$(".zetin16").text(value);
} else {
alert('this is bad');
}
});
But I get this error: Uncaught SyntaxError: Invalid or unexpected token
In this line: var regex = "[A-Z0-9a-z&,.-/()#*+!?"':; -]";
Thanks in advance for any help.
UPDATE
The regex working fine now. Now I want to prevent typing characters in input when regex doesnt match the character. This is my code currently:
$("#firstText").keyup(function(e) {
var value = $(this).val().toUpperCase();
var regex = new RegExp(/[A-Z0-9a-z&,.-/()#*+!?"':; -]/);
if (regex.test(value)) {
$(".zetin16").text(value);
} else {
e.preventDefault();
return false;
}
});

With regex, use the forward slash as delimiter. If a forward slash occurs as a literal inside the regex itself, escape it:
var regex = /[A-Z0-9a-z&,.-\/()#*+!?"':; -]/;
Reference: JavaScript regex replace - escaping slashes
(The problem with the original string was that it contained a double quote, and was delimited using double quotes at the same time).

The exact error you're seeing is because you are defining the variable as a double quoted string, with an unescaped double quote in it.
It shouldn't be a string anyway. It should be a regular expression like this.
var regex = /[A-Z0-9a-z&,.-/()#*+!?"':; -]/;

try using this pattern for using regular expression
var regex = "['A-Z0-9a-z&,.-/()#*+!?':; -]";
var reg =new RegExp(regex)
var val ="asss"
reg.test(val)

Regex: find capitalized words [closed]

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How do I find and extract capitalized words of a string with regex?
I would like to:
extract the capitalized words of a string, as an array
extract the last capitalized word of a string, as a substring:
Both with one regex
If I have this:
var str="This is a STRING of WORDS to search";
I would like to get this 1:
allCapWords // = ["STRING", "WORDS"]
and 2:
lastCapWord // = "WORDS"

To extract the words into an array:
var allCapWords = str.match(/\b[A-Z]+\b/g);
-> ["STRING", "WORDS"]
(Here's a Regex101 test with your string.)
To pull the last word:
var lastCapWord = allCapWords[allCapWords.length - 1];
-> "WORDS"

var str="This is a STRING of WORDS to search";
var regObj = /\b([A-Z]+)\b/g;
allCapWords = str.match(regObj);

You can try this regexpr /\b[A-Z]+\b/gor \b[A-Z0-9]+\b/g if you are interested in catch numbers inside the string

Get part of the string using regexp [closed]

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I've a strings, which can have a text like:
'some text user#t12# some text'
'username#John# some text'
'some text usersurname#Malks#'
'userphoto#1.jpg#'
How do I get a text between # and # symbols?
There's a typical structure of the part of the string to search for - type#variable#
type is a JS variable type, it's placed before the first #.
variable is a text that I need to get.
I'm searching for a regexp, that return variable, that is between #...#.
The problem is, I'm not too familiar with regexp, can you help me please?

You need to use capture groups, basically in a regex anything in brackets will be part of the cpature group, in this case you want to capture all the characters between two hashes. The any amount of characters regex is .* so this is what you want to capture between two hashes. Once you execute it you will find the match as second in the array (the first will be the string with the hashes.
var type = "";
var myString = "some text user#t12# some text";
var myRegexp = new RegExp(type+"#(.*)#","g");
var match = myRegexp.exec(myString);
alert(match[1]); // t12
any other matches between hashes will be in match[2].. match[n]