Develop Reference

Javascript Regex - replacing characters based on regex rules

I am trying to remove illegal characters from a user input on a browser input field.
const myInput = '46432e66Sc'
var myPattern = new RegExp(/^[a-z][a-z0-9]*/);
var test = myPattern.test(myInput);
if (test === true) {
console.log('success',myInput)
} else {
console.log("fail",myInput.replace(???, ""))
}
I can test with the right regex and it works just fine. Now I am trying to remove the illegal characters. The rules are, only lower case alpha character in the first position. All remaining positions can only have lower case alpha and numbers 0-9. No spaces or special characters. I am not sure what pattern to use on the replace line.
Thanks for any help you can provide.
Brad

You could try the below code:
const myInput = '46432e66Sc'
var myPattern = new RegExp(/^[a-z][a-z0-9]*/);
var test = myPattern.test(myInput);
if (test === true) {
console.log('success',myInput)
} else {
console.log("fail",myInput.replace(/[^a-z0-9]/g, ""))
}
Replace is using the following regexp: /[^a-z0-9]/g. This matches all characters that are not lowercase or numeric.
You can validate your regexp and get help from cheatsheet on the following page: https://regexr.com/

You could handle this by first stripping off any leading characters which would cause the input to fail. Then do a second cleanup on the remaining characters:
var inputs = ['abc123', '46432e66Sc'];
inputs.forEach(i => console.log(i + " => " + i.replace(/^[^a-z]+/, "")
.replace(/[^a-z0-9]+/g, "")));
Note that after we have stripped off as many characters as necessary for the input to start with a lowercase, the replacement to remove non lowercase/non digits won't affect that first character, so we can just do a blanket replacement on the entire string.

Only allow whitespace after comma and restrict special characters in text field using regex

I am trying to restrict a user from being able to enter whitespace or hit the spacebar after characters (unless they enter a comma) as well as restrict all special characters except numbers and letters.
The desired output would be something like: "ab, c" or "abc, def, g, h..."
The user would only be allowed to have whitespace or hit the spacebar after a comma. A comma is the only special character permitted.
What I have so far but it's only allowing for 1 comma:
function checkKey() {
var clean = this.value.replace(/[^0-9,]/g, "")
.replace(/(,.*?),(.*,)?/, "$1");
// don't move cursor to end if no change
if (clean !== this.value) this.value = clean;
}
// demo
document.querySelector('input').oninput = checkKey;

In your code .replace(/[^0-9,]/g, "") , you are removing all chars except digits and a comma, not allowing chars a-z anymore.
You can first remove all characters except [^a-zA-Z0-9, ]+, and then in browsers that support a lookbehind, you can allow a space when it is not directly preceded by a comma.
function checkKey() {
const clean = this.value
.replace(/[^a-zA-Z0-9, ]+/g, "")
.replace(/(?<!,) /g, "");
if (clean !== this.value) this.value = clean;
}
document.querySelector('input').oninput = checkKey;
<form>
<input type="text">
</form>

To remove all whitespaces after a non-comma char, you can use this solution with no lookbehind:
.replace(/(^|[^,])\s+/g, "$1")
See this regex demo. Details:
(^|[^,]) - Capturing group 1 ($1 refers to this group value): start of string (^) or (|) any non-comma char ([^,])
\s+ - one or more whitespaces.
In your code:
function checkKey() {
var clean = this.value.replace(/[^\d,]/g, "").replace(/(^|[^,])\s+/g, "$1");
if (clean !== this.value) {
this.value = clean;
}
}

Javascript remove all characters by regex rules

Who can help me with the following
I create a rule with regex and I want remove all characters from the string if they not allowed.
I tried something by myself but I get not the result that I want
document.getElementById('item_price').onkeydown = function() {
var regex = /^(\d+[,]+\d{2})$/;
if (regex.test(this.value) == false ) {
this.value = this.value.replace(regex, "");
}
}
The characters that allowed are numbers and one komma.
Remove all letters, special characters and double kommas.
If the user types k12.40 the code must replace this string to 1240
Who can help me to the right direction?

This completely removes double occurrences of commas using regex, but keeps single ones.
// This should end up as 1,23243,09
let test = 'k1,23.2,,43d,0.9';
let replaced = test.replace(/([^(\d|,)]|,{2})/g, '')
console.log(replaced);

I don't believe there's an easy way to have a single Regex behave like you want. You can use a function to determine what to replace each character with, though:
// This should end up as 1232,4309 - allows one comma and any digits
let test = 'k12,3.2,,43,d0.9';
let foundComma = false;
let replaced = test.replace(/(,,)|[^\d]/g, function (item) {
if (item === ',' && !foundComma) {
foundComma = true;
return ',';
} else {
return '';
}
})
console.log(replaced);
This will loop through each non-digit. If its the first time a comma has appeared in this string, it will leave it. Otherwise, if it must be either another comma or a non-digit, and it will be replaced. It will also replace any double commas with nothing, even if it is the first set of commas - if you want it to be replaced with a single comma, you can remove the (,,) from the regex.

Replace after char '-' or '/' match

I'm trying to execute regex replace after match char, example 3674802/3 or 637884-ORG
The id can become one of them, in that case, how can I use regex replace to match to remove after the match?
Input var id = 3674802/3 or 637884-ORG;
Expected Output 3674802 or 637884

You could use sbustring method to take part of string only till '/' OR '-':
var input = "3674802/3";
var output = input.substr(0, input.indexOf('/'));
var input = "637884-ORG";
var output = input.substr(0, input.indexOf('-'));
var input = "3674802/3";
if (input.indexOf('/') > -1)
{
input = input.substr(0, input.indexOf('/'));
}
console.log(input);
var input = "637884-ORG";
if (input.indexOf('-') > -1)
{
input = input.substr(0, input.indexOf('-'));
}
console.log(input);

You can use a regex with a lookahead assertion
/(\d+)(?=[/-])/g
var id = "3674802/3"
console.log((id.match(/(\d+)(?=[/-])/g) || []).pop())
id = "637884-ORG"
console.log((id.match(/(\d+)(?=[/-])/g) || []).pop())

You don't need Regex for this. Regex is far more powerful than what you need.
You get away with the String's substring and indexOf methods.
indexOf takes in a character/substring and returns an integer. The integer represents what character position the character/substring starts at.
substring takes in a starting position and ending position, and returns the new string from the start to the end.
If are having trouble getting these to work; then, feel free to ask for more clarification.

You can use the following script:
var str = '3674802/3 or 637884-ORG';
var id = str.replace(/(\d+)[-\/](?:\d+|[A-Z]+)/g, '$1');
Details concerning the regex:
(\d+) - A seuence of digits, the 1st capturing group.
[-\/] - Either a minus or a slash. Because / are regex delimiters,
it must be escaped with a backslash.
(?: - Start of a non-capturing group, a "container" for alternatives.
\d+ - First alternative - a sequence of digits.
| - Alternative separator.
[A-Z]+ - Second alternative - a sequence of letters.
) - End of the non-capturing group.
g - global option.
The expression to replace with: $1 - replace the whole finding with
the first capturing group.

Thanks To everyone who responded to my question, was really helpful to resolve my issue.
Here is My answer that I built:
var str = ['8484683*ORG','7488575/2','647658-ORG'];
for(i=0;i<str.length;i++){
var regRep = /((\/\/[^\/]+)?\/.*)|(\-.*)|(\*.*)/;
var txt = str[i].replace(regRep,"");
console.log(txt);
}

javascript regular expression test for 6 digit numbers only. comma seperated

and so this must pass:
454555, 939999 , 019999 ,727663
its for a user entering 6 digit invoice numbers. it should fail if a number is 5 or 7 digit and not 6. so 1234567, 123456 should fail, as one set is more than 6 numbers.
So far I have :
[0-9]{6}(\s*,*,\s*[0-9]{6})*
which only draw back is that it accepts 7 or more digit numbers. cant figure out if its even possible at this point to do both, test for 6 digits separated by a comma and one or more space, and all the digits have to be only 6 digits and fail if one is not.
any help appreciated. regular expressions are not my forte.
thanks
Norman

You can write it using regex like the function below.
const isPassword = (password: string) => /^\d{6}$/gm.test(password);
And here is an example test file below.
test('should recognize a valid password', () => {
expect(isPassword('123456')).toBe(true);
expect(isPassword('000000')).toBe(true);
});
test('should recognize an invalid password', () => {
expect(isPassword('asdasda1234')).toBe(false);
expect(isPassword('1234567')).toBe(false);
expect(isPassword('a123456a')).toBe(false);
expect(isPassword('11.11.11')).toBe(false);
expect(isPassword('aaaaaa')).toBe(false);
expect(isPassword('eeeeee')).toBe(false);
expect(isPassword('......')).toBe(false);
expect(isPassword('werwerwerwr')).toBe(false);
});

In order to validate the full string you can use this regex.
^(\s*\d{6}\s*)(,\s*\d{6}\s*)*,?\s*$
It works with six digits only, and you have to enter at least one 6 digit number.
It also works if you have a trailing comma with whitespaces.

It's accepting more than six digit numbers because you're not anchoring the text, and for some odd reason you're optionally repeating the comma. Try something like this:
^[0-9]{6}(?:\s*,\s*[0-9]{6})*$
Also note that [0-9] is equivalent to \d, so this can be rewritten more concisely as:
^\d{6}(?:\s*,\s*\d{6})*$

Your regex does not match 7 digits in a row, but it also doesn't enforce that it matches the whole string. It just has to match some substring in the string, so it would also match each of these:
"1234512345612345612345"
"NaNaNaN 123456, 123456 BOOO!"
"!##$%^&*({123456})*&^%$##!"
Just add the start of string (^) and end of string ($) anchors to enforce that the whole string matches and it will work correctly:
^[0-9]{6}(\s*,*,\s*[0-9]{6})*$
Also note that ,*, could be shortened to ,+, and if you only want one comma in a row, just use ,, not ,* or ,+.
You can also replace [0-9] with \d:
^\d{6}(\s*,\s*\d{6})*$

Using only regex:
var commaSeparatedSixDigits = /^(?:\d{6}\s*,\s*)*\d{6}$/;
if (myInput.test(commaSeparatedSixDigits)) console.log( "Is good!" );
This says:
^ - Starting at the beginning of the string
(?:…)* - Find zero or more of the following:
\d{6} - six digits
\s* - maybe some whitespace
, - a literal comma
\s* - maybe some whitespace
\d{6} - Followed by six digits
$ - Followed by the end of the string
Alternatively:
var commaSeparatedSixDigits = /^\s*\d{6}(?:\s*,\s*\d{6})*\s*$/;
I leave it as an exercise to you to decipher what's different about this.
Using JavaScript + regex:
function isOnlyCommaSeparatedSixDigitNumbers( str ){
var parts = srt.split(/\s*,\s*/);
for (var i=parts.length;i--;){
// Ensure that each part is exactly six digit characters
if (! /^\d{6}$/.test(parts[i])) return false;
}
return true;
}

I see a lot of complication here. Sounds to me like what you want is pretty simple:
/^(\d{6},)*\d{6}$/
Then we account for whitespace:
/^\s*(\d{6}\s*,\s*)*\d{6}\s*$/
But as others have noted, this is actually quite simple in JavaScript without using regex:
function check(input) {
var parts = input.split(',');
for (var i = 0, n = parts.length; i < n; i++) {
if (isNaN(+parts[i].trim())) {
return false;
}
}
return true;
}
Tested in the Chrome JavaScript console.

There isn;t any real need for a regexp. Limit the input to only 6 characters, only accept numbers and ensure that the input has 6 digits (not show here). So you would need:
HTML
<input type='text' name='invoice' size='10' maxlength='6' value='' onkeypress='evNumersOnly(event);'>
JavaScript
<script>
function evNumbersOnly( evt ) {
//--- only accepts numbers
//--- this handles incompatabilities between browsers
var theEvent = evt || window.event;
//--- this handles incompatabilities between browsers
var key = theEvent.keyCode || theEvent.which;
//--- convert key number to a letter
key = String.fromCharCode( key );
var regex = /[0-9]/; // Allowable characters 0-9.+-,
if( !regex.test(key) ) {
theEvent.returnValue = false;
//--- this prevents the character from being displayed
if(theEvent.preventDefault) theEvent.preventDefault();
}
}
</script>