Develop Reference

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.

Regex to validate a texarea input which must be URLs separated by new lines

I am trying to create a regex which will ultimately be used with Google Forms to validate a texarea input.
The rule is,
Input area can have one or more URLs (http or https)
Each URL must be separated either by one or more new lines
Each line which has text, must be a single valid URL
Last URL may have or may not have new line character/s after it
Till now, I have written this regex ^(https?://.+[\r\n]+)*(https?://.+[\r\n]+?)$ but the problem is that if a line has more than 1 url, it validates that too.
Here is my testing playground: http://goo.gl/YPdvBH.

Here is what you are looking for
Demo , Demo with your URLS
function validate(ele) {
str = ele.value;
str = str.replace(/\r/g, "");
while (/\s\n/.test(str)) {
str = str.replace(/\s\n/g, "\n");
}
while (/\n\n/.test(str)) {
str = str.replace(/\n\n/g, "\n");
}
ele.value = str;
str = str.replace(/\n/g, "_!_&_!_").split("_!_&_!_")
var result = [], counter = 0;
for (var i = 0; i < str.length; i++) {
str[i] = str[i].replace(/(?:(?:^|\n)\s+|\s+(?:$|\n))/g, '').replace(/\s+/g, ' ');
if(str[i].length !== 0){
if (isValidAddress(str[i])) {
result.push(str[i]);
}
counter += 1;
}
}
function isValidAddress(s) {
return /^(https?|ftp):\/\/(((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:)*#)?(((\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\.(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\.(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5])\.(\d|[1-9]\d|1\d\d|2[0-4]\d|25[0-5]))|((([a-z]|\d|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(([a-z]|\d|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])*([a-z]|\d|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])))\.)+(([a-z]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(([a-z]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])*([a-z]|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])))\.?)(:\d*)?)(\/((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|#)+(\/(([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|#)*)*)?)?(\?((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|#)|[\uE000-\uF8FF]|\/|\?)*)?(\#((([a-z]|\d|-|\.|_|~|[\u00A0-\uD7FF\uF900-\uFDCF\uFDF0-\uFFEF])|(%[\da-f]{2})|[!\$&'\(\)\*\+,;=]|:|#)|\/|\?)*)?$/i.test(s)
}
return (result.length === str.length);
}
var ele = document.getElementById('urls');
validate(ele);

This is closer to the regex you are looking for:
^(https?://[\S]+[\r\n]+)*(https?://[\S]+[\r\n]+?)$
The difference between your regex and this one is that you use .+ which will match all characters except newline whereas I use [\S]+ (note it is a capital S) which will match all non-whitespace characters. So, this doesn't match more than one token on one line. Hence, on each line you can match at max one token and that must be of the form that you have defined.
For a regex to match a single URL, look at this question on StackOverflow:
What is the best regular expression to check if a string is a valid URL?
I don't know whether google-forms have a length limit. But if they have, it is sure to almost bounce into it.

If i understand right - in your regexp missing m flag for multiline, so you need something like this
/^(https?://.+this your reg exp for one url)$/m
sample with regexp from Javascript URL validation regex
/^(ht|f)tps?:\/\/[a-z0-9-\.]+\.[a-z]{2,4}\/?([^\s<>\#%"\,\{\}\\|\\\^\[\]`]+)?$/m

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>

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

How to check for uppercase alphabets in an input string, using jQuery

I am using following code snippet, but its not working :-(
//First four characters of input Text should be ALPHABATES (Letters)
if (($("#txtId").val()).length >= 4) {
var firstFourChars = $("#txtId").val().substring(0, 4);
var pattern = new RegExp('[^A-Z]');
if (firstFourChars.match(pattern))
isValid = true;
else
isValid = false;
}

change /[^A-Z]/ to /^[A-Z]/
example :
var a = "ABCJabcd";
console.log(a.match(/^[A-Z]{4}/));

you don't need to use substring(). Your regexp can do all the work for you. The RegExp you are using matches against characters that are NOT between A and Z. As Avinash said, ^[A-Z]{4} will match if your first 4 characters are uppercase. "^" at the beginning of your regexp tells that the following should be the beginning of the string. When placed inside square brackets, it reverts the range of characters you want to match.

The regex should be /[^A-Z]{4}/ if you want to match the 4 lowercase characters.

To detect in the middle of the big papers change /^[A-Z]/ to /[A-Z]/
Example text: " asşldla ABCJ abcd AÇALASD"
$('.Order input').change(function (){ucheck($(this).val())});
$('.Order input').keyup(function (){ucheck($(this).val())});
function ucheck(a) {
if(a.match(/[A-ZĞÜŞİÖÇ]{4}/)){
$('.Order #Error').html(' UPPERCASE');
}else{$('.Order #Error').html('Capitalize');}
}

If they need to be capital:
const startsWithCapitals = /^[A-Z]{4}/.test(string);
Or if they just need to be letters, add an i for ignore case:
const startsWithLetters = /^[a-z]{4}/i.test(string);
^ means start of the string and {number} means x copies