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.

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.

Detect repeating letter in an string in Javascript

code for detecting repeating letter in a string.
var str="paraven4sr";
var hasDuplicates = (/([a-zA-Z])\1+$/).test(str)
alert("repeating string "+hasDuplicates);
I am getting "false" as output for the above string "paraven4sr". But this code works correctly for the strings like "paraaven4sr". i mean if the character repeated consecutively, code gives output as "TRUE". how to rewrite this code so that i ll get output as "TRUE" when the character repeats in a string

JSFIDDLE
var str="paraven4sr";
var hasDuplicates = (/([a-zA-Z]).*?\1/).test(str)
alert("repeating string "+hasDuplicates);
The regular expression /([a-zA-Z])\1+$/ is looking for:
([a-zA-Z]]) - A letter which it captures in the first group; then
\1+ - immediately following it one or more copies of that letter; then
$ - the end of the string.
Changing it to /([a-zA-Z]).*?\1/ instead searches for:
([a-zA-Z]) - A letter which it captures in the first group; then
.*? - zero or more characters (the ? denotes as few as possible); until
\1 - it finds a repeat of the first matched character.
If you have a requirement that the second match must be at the end-of-the-string then you can add $ to the end of the regular expression but from your text description of what you wanted then this did not seem to be necessary.

Try this:
var str = "paraven4sr";
function checkDuplicate(str){
for(var i = 0; i < str.length; i++){
var re = new RegExp("[^"+ str[i] +"]","g");
if(str.replace(re, "").length >= 2){
return true;
}
}
return false;
}
alert(checkDuplicate(str));
Here is jsfiddle

To just test duplicate alphanumeric character (including underscore _):
console.log(/(\w)\1+/.test('aab'));

Something like this?
String.prototype.count=function(s1) {
return (this.length - this.replace(new RegExp(s1,"g"), '').length) / s1.length;
}
"aab".count("a") > 1
EDIT: Sorry, just read that you are not searching for a function to find whether a letter is found more than once but to find whether a letter is a duplicate. Anyway, I leave this function here, maybe it can help. Sorry ;)

How to match one, but not two characters using regular expressions

Using javascript regular expressions, how do you match one character while ignoring any other characters that also match?
Example 1: I want to match $, but not $$ or $$$.
Example 2: I want to match $$, but not $$$.
A typical string that is being tested is, "$ $$ $$$ asian italian"
From a user experience perspective, the user selects, or deselects, a checkbox whose value matches tags found in in a list of items. All the tags must be matched (checked) for the item to show.
function filterResults(){
// Make an array of the checked inputs
var aInputs = $('.listings-inputs input:checked').toArray();
// alert(aInputs);
// Turn that array into a new array made from each items value.
var aValues = $.map(aInputs, function(i){
// alert($(i).val());
return $(i).val();
});
// alert(aValues);
// Create new variable, set the value to the joined array set to lower case.
// Use this variable as the string to test
var sValues = aValues.join(' ').toLowerCase();
// alert(sValues);
// sValues = sValues.replace(/\$/ig,'\\$');
// alert(sValues);
// this examines each the '.tags' of each item
$('.listings .tags').each(function(){
var sTags = $(this).text();
// alert(sTags);
sSplitTags = sTags.split(' \267 '); // JavaScript uses octal encoding for special characters
// alert(sSplitTags);
// sSplitTags = sTags.split(' \u00B7 '); // This also works
var show = true;
$.each(sSplitTags, function(i,tag){
if(tag.charAt(0) == '$'){
// alert(tag);
// alert('It begins with a $');
// You have to escape special characters for the RegEx
tag = tag.replace(/\$/ig,'\\$');
// alert(tag);
}
tag = '\\b' + tag + '\\b';
var re = new RegExp(tag,'i');
if(!(re.test(sValues))){
alert(tag);
show = false;
alert('no match');
return false;
}
else{
alert(tag);
show = true;
alert('match');
}
});
if(show == false){
$(this).parent().hide();
}
else{
$(this).parent().show();
}
});
// call the swizzleRows function in the listings.js
swizzleList();
}
Thanks in advance!

Normally, with regex, you can use (?<!x)x(?!x) to match an x that is not preceded nor followed with x.
With the modern ECMAScript 2018+ compliant JS engines, you may use lookbehind based regex:
(?<!\$)\$(?!\$)
See the JS demo (run it in supported browsers only, their number is growing, check the list here):
const str ="$ $$ $$$ asian italian";
const regex = /(?<!\$)\$(?!\$)/g;
console.log( str.match(regex).length ); // Count the single $ occurrences
console.log( str.replace(regex, '<span>$&</span>') ); // Enclose single $ occurrences with tags
console.log( str.split(regex) ); // Split with single $ occurrences

\bx\b
Explanation: Matches x between two word boundaries (for more on word boundaries, look at this tutorial). \b includes the start or end of the string.
I'm taking advantage of the space delimiting in your question. If that is not there, then you will need a more complex expression like (^x$|^x[^x]|[^x]x[^x]|[^x]x$) to match different positions possibly at the start and/or end of the string. This would limit it to single character matching, whereas the first pattern matches entire tokens.
The alternative is just to tokenize the string (split it at spaces) and construct an object from the tokens which you can just look up to see if a given string matched one of the tokens. This should be much faster per-lookup than regex.

Something like that:
q=re.match(r"""(x{2})($|[^x])""", 'xx')
q.groups() ('xx', '')
q=re.match(r"""(x{2})($|[^x])""", 'xxx')
q is None True

Remove all characters after last - but only if after - has four characters

I have a string which could contain several different values, among them are.
EDITED for clarity:
var test could equal FW21002-185 or FW21002-181-0001 or abcdefg or 245-453-654 or FW21002-181-00012
I would like to remove all characters after and including the last - only if that string contains four characters after the last dash. So in the above strings examples, the only one that should be changed is the second one to "FW21002-181" All others would remain as they are.
How would I do this in JavaScript. Regex is ok as well. Thanks.

A regex to do this would be
var chopped = test.replace(/-[^-]{4,}$/, '-');
(assuming you want that "-" at the end). (Oh also this is intended to match 4 or more trailing characters - if you want exactly four, just get rid of the comma in {4,}.)

No regex required:
var str = ...,
pos = str.lastIndexOf('-');
if (pos > -1 && pos == str.length - 5)
str = str.substring(0, pos);

If you don't want to use a regex:
function removeLongSuffix(var str)
{
var tokens = str.split('-'),
last = tokens[tokens.length-1];
if (last.length > 3)
{
return tokens.slice(0,-1).join('-');
}
return str;
}