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Extract number and text from string with RegExp exec Javascript

I'm trying to extract a number and text from strings like those: 171Toberin, [171]Toberin or [171] Toberin.
I have this RegExp /(?<code>\d+)(?<name>\w+)/u, and this RegExp only works with 171Toberin.

You can use the below regex to remove all non alphanumeric characters.
let string = '171Toberin, [171]Toberin or [171]';
console.log(string.replace(/[^a-zA-Z0-9]/g, ''));
Or use the below to extract the alphanumeric characters from string.
let string = '171Toberin, [171]Toberin or [171]';
console.log(string.match(/[a-zA-Z0-9]+/g));
Or if you want to extract numbers and strings in separate array then use the below one.
let string = '171Toberin, [171]Toberin or [171]';
console.log(string.match(/[0-9]+/g));
console.log(string.match(/[a-zA-Z]+/g));

Please try with this: (?<code>\d+)[^\d\w]*(?<name>\w+)/ug
It works with the entire sentence: 171Toberin, [171]Toberin or [171] Toberin.
Returning 3 matches. You can try it at https://regex101.com/
With [^\d\w]* you omit possible numbers and words in between. With g flag you return all matches.

how to check comma between two string and build a regex for that

I have a text box and I need to have validation that no commas allowed between two words like this given below eg B521,Baraghat. I want to have regext for that. How to do that in javascript.

A regex is not necessary for this check. This will make sure that your string does not contain a comma.
if(str.indexOf(',') != -1) {
<Invalid string logic>
}
If you insist on a regex:
if(str.match(/\,/)) {
<invalid string logic>
}

var patt=/[^,]+,[^,]+/
That will find a comma between two non-commas.
This matches both:
B521,Baraghat
and
B521 , Baraghat
If you want to find a comma between two words explicitly, you can do:
var patt=/\b,\b/
This will find a comma between two word boundaries.
This matches
B521,Baraghat
but not
B521 , Baraghat

Matching special characters and letters in regex

I am trying to validate a string, that should contain letters numbers and special characters &-._ only. For that I tried with a regular expression.
var pattern = /[a-zA-Z0-9&_\.-]/
var qry = 'abc&*';
if(qry.match(pattern)) {
alert('valid');
}
else{
alert('invalid');
}
While using the above code, the string abc&* is valid. But my requirement is to show this invalid. ie Whenever a character other than a letter, a number or special characters &-._ comes, the string should evaluate as invalid. How can I do that with a regex?

Add them to the allowed characters, but you'll need to escape some of them, such as -]/\
var pattern = /^[a-zA-Z0-9!##$%^&*()_+\-=\[\]{};':"\\|,.<>\/?]*$/
That way you can remove any individual character you want to disallow.
Also, you want to include the start and end of string placemarkers ^ and $
Update:
As elclanrs understood (and the rest of us didn't, initially), the only special characters needing to be allowed in the pattern are &-._
/^[\w&.\-]+$/
[\w] is the same as [a-zA-Z0-9_]
Though the dash doesn't need escaping when it's at the start or end of the list, I prefer to do it in case other characters are added. Additionally, the + means you need at least one of the listed characters. If zero is ok (ie an empty value), then replace it with a * instead:
/^[\w&.\-]*$/

Well, why not just add them to your existing character class?
var pattern = /[a-zA-Z0-9&._-]/
If you need to check whether a string consists of nothing but those characters you have to anchor the expression as well:
var pattern = /^[a-zA-Z0-9&._-]+$/
The added ^ and $ match the beginning and end of the string respectively.
Testing for letters, numbers or underscore can be done with \w which shortens your expression:
var pattern = /^[\w&.-]+$/
As mentioned in the comment from Nathan, if you're not using the results from .match() (it returns an array with what has been matched), it's better to use RegExp.test() which returns a simple boolean:
if (pattern.test(qry)) {
// qry is non-empty and only contains letters, numbers or special characters.
}
Update 2
In case I have misread the question, the below will check if all three separate conditions are met.
if (/[a-zA-Z]/.test(qry) && /[0-9]/.test(qry) && /[&._-]/.test(qry)) {
// qry contains at least one letter, one number and one special character
}

Try this regex:
/^[\w&.-]+$/
Also you can use test.
if ( pattern.test( qry ) ) {
// valid
}

let pattern = /^(?=.*[0-9])(?=.*[!##$%^&*])(?=.*[a-z])(?=.*[A-Z])[a-zA-Z0-9!##$%^&*]{6,16}$/;
//following will give you the result as true(if the password contains Capital, small letter, number and special character) or false based on the string format
let reee =pattern .test("helLo123#"); //true as it contains all the above

I tried a bunch of these but none of them worked for all of my tests. So I found this:
^(?=.*\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[^a-zA-Z0-9])(?!.*\s).{8,15}$
from this source: https://www.w3resource.com/javascript/form/password-validation.php

Try this RegEx: Matching special charecters which we use in paragraphs and alphabets
Javascript : /^[a-zA-Z]+(([\'\,\.\-_ \/)(:][a-zA-Z_ ])?[a-zA-Z_ .]*)*$/.test(str)
.test(str) returns boolean value if matched true and not matched false
c# : ^[a-zA-Z]+(([\'\,\.\-_ \/)(:][a-zA-Z_ ])?[a-zA-Z_ .]*)*$

Here you can match with special char:
function containsSpecialChars(str) {
const specialChars = /[`!##$%^&*()_+\-=\[\]{};':"\\|,.<>\/?~]/;
return specialChars.test(str);
}
console.log(containsSpecialChars('hello!')); // 👉️ true
console.log(containsSpecialChars('abc')); // 👉️ false
console.log(containsSpecialChars('one two')); // 👉️ false

How to make this simple regexp?

I need to make a string starts and ends with alphanumeric range between 5 to 20 characters and it could have a space or none between characters. /^[a-z\s?A-Z0-9]{5,20}$/ but this is not working.
EDIT
test test -should pass
testtest -should pass
test test test -should not pass

You can't do this with traditional regex without writing a ridiculously long expression, so you need to use a look-ahead:
/^(?=(\w| ){15,20}$)\w+ ?\w+$/
This says, make sure there are between 15 and 20 characters in the match, then match /\w+ \w+/
Note I used \w for simplification. It is the same as your character class above except it also accepts underscores. If you don't want to match them you have to do:
/^(?=[a-zA-Z0-9 ]{15,20}$)[a-zA-Z0-9]+ ?[a-zA-Z0-9]+$/

You can't put a ? inside of [...]. [...] is used to specify a set of characters precisely, you can't maybe (?) have a character inside a set of characters. The occurrence of any specific characters is already optional, the ? is meaningless.
If you allow any number of spaces inside your match, just remove the question mark. If you want to allow a single space but no more, then regular expressions alone can't do that for you, you'd need something like
if (myString.match(/^[a-z\sA-Z0-9]{5,20}$/ && myString.match(/\s/g).length <= 1)
You couldn't do this with a single traditional regex without it being dozens of lines long; regexes are meant for matching more simpler patterns than this.
If you only want to use regexes, you could use two instead of one. The first matches the general pattern, the second ensures that only one non-space characters is found.
if (myString.match(/^[a-z\sA-Z0-9]{5,20}$/ && myString.match(/^[^\s]*\s?[^\s]*$/))) {
Example Usage
inputs = ["test test", "testtest", "test test test"];
for (index in inputs) {
var myString = inputs[index];
if (myString.match(/^[a-z\sA-Z0-9]{5,20}$/ && myString.match(/^[^\s]*\s?[^\s]*$/))) {
console.log(myString + " matches.")
} else {
console.log(myString + " does not match.")
}
}
This produces the output specified in your question.

Meh , So here's the ridiculously long traditional regex for the same
(?i)[a-z0-9]+( [a-z0-9]+)?{5,12}
js vesrion (w/o the nested quantifier)
/^([a-z0-9]( [a-z0-9])?){5,12}$/i

Javascript Regex: replacing the last dot for a comma

I have the following code:
var x = "100.007"
x = String(parseFloat(x).toFixed(2));
return x
=> 100.01
This works awesomely just how I want it to work. I just want a tiny addition, which is something like:
var x = "100,007"
x.replace(",", ".")
x.replace
x = String(parseFloat(x).toFixed(2));
x.replace(".", ",")
return x
=> 100,01
However, this code will replace the first occurrence of the ",", where I want to catch the last one. Any help would be appreciated.

You can do it with a regular expression:
x = x.replace(/,([^,]*)$/, ".$1");
That regular expression matches a comma followed by any amount of text not including a comma. The replacement string is just a period followed by whatever it was that came after the original last comma. Other commas preceding it in the string won't be affected.
Now, if you're really converting numbers formatted in "European style" (for lack of a better term), you're also going to need to worry about the "." characters in places where a "U.S. style" number would have commas. I think you would probably just want to get rid of them:
x = x.replace(/\./g, '');
When you use the ".replace()" function on a string, you should understand that it returns the modified string. It does not modify the original string, however, so a statement like:
x.replace(/something/, "something else");
has no effect on the value of "x".

You can use a regexp. You want to replace the last ',', so the basic idea is to replace the ',' for which there's no ',' after.
x.replace(/,([^,]*)$/, ".$1");
Will return what you want :-).

You could do it using the lastIndexOf() function to find the last occurrence of the , and replace it.
The alternative is to use a regular expression with the end of line marker:
myOldString.replace(/,([^,]*)$/, ".$1");

You can use lastIndexOf to find the last occurence of ,. Then you can use slice to put the part before and after the , together with a . inbetween.

You don't need to worry about whether or not it's the last ".", because there is only one. JavaScript doesn't store numbers internally with comma or dot-delimited sets.