I've been trying for ages now using JavaScript to get a regex that does not allow symbols however allows hyphens and underscores but it just isn't working, here is what I have tried:
function checkUsername(username)
{
username_lbl = document.getElementById('<%= username_lbl.ClientID %>');
register_btn = document.getElementById('<%= register_btn.ClientID %>');
if (username.match("?[!##$%^&*()+=[{]};:<>|./?.]"))
{
username_lbl.innerHTML = 'Cannot contain symbols';
register_btn.disabled = true;
username_lbl.style.color = "red";
}
}
Does anyone understand what I am doing wrong?
Because many of the symbols (e.g. ?, [,] ) are used as a regex symbol. You need to either escape them, or do this:
username.match("[^\\w-_]");
[^] means does not match character set.
\w means word character [a-zA-Z0-9]
EdBallot Edit: Fixed typo regex (was [^\\w-_], changed to [^\w\-_])
Author Edit: It should be "[^\\w-_]" because it is double quoted and should be escaped, or you can single quote or make it regex
username.match(/[^\w-_]/)
Just use
var username = "_Fred-Fred_";
if (/[^-_a-zA-Z]/.test(username)) {
alert("Wrong username: " + username);
}
username = "!_Fred-Fred_";
if (/[^-_a-zA-Z]/.test(username)) {
alert("Wrong username: " + username);
}
The [^-_a-zA-Z] matches anything but a letter, a hyphen or an underscore. If something else is found, the test is failed.
var input = [paul, Paula, george];
var newReg = \paula?\i
for(var text in input) {
if (newReg.test(text) == true) {
input[input.indexOf(text)] = george
}
}
console.log(input)
I don't know what's wrong in my code. it should change paul and Paula to george but when I run it it says there's an illegal character
The backslash (\) is an escape character in Javascript (along with a lot of other C-like languages). This means that when Javascript encounters a backslash, it tries to escape the following character. For instance, \n is a newline character (rather than a backslash followed by the letter n).
So, thats what is causing your error, you need to replace \paula?\i with /paula?/i
You need to replace \ by / in your regexp pattern.
You should wrap the strings inside quotes "
You need to match correctly your array, val is just the index of the word, not the word himself.
var input = ["paul", "Paula", "george"];
var newReg = /paula?/i;
for (var val in input) {
if (newReg.test(input[val]) == true) {
input[input.indexOf(input[val])] = "george";
}
}
console.log(input);
JSFIDDLE
I have a function which validates the phone number entered by the user using regexp. However it does not seem to evaluate true even though the regexp is correct. I am not sure what i am doing incorrectly.
Html
<body class="claro">
<form id="myform" data-dojo-type="dijit/form/Form">
<input
data-dojo-type="dijit/form/ValidationTextBox"
data-dojo-props="
required: true,
invalidMessage: 'Invalid Phone Number !',
missingMessage: 'Phone Number Required !'"
id="phone" title="Phone Number"
placeholder="Your Phone Number"
onkeydown="validatePhoneNumberFormat()"/>
</form>
</body>
Javascript
//test phone number 188-123-1234
function validatePhoneNumberFormat(){
var phoneNumber = dijit.byId("phone");
var phoneFormat = new RegExp('^[0-9]\d{2}-\d{3}-\d{4}$');
phoneNumber.validator = function(value){
console.log(value);
console.log(phoneFormat.test(value.trim()));
return phoneFormat.test(value.trim());
}
}
You need to double escape \d inside RegExp constructor, so use this:
var phoneFormat = new RegExp('^\\d{3}-\\d{3}-\\d{4}$');
Or else use regex literal:
var phoneFormat = /^\d{3}-\d{3}-\d{4}$/;
Since RegExp takes a string as an argument you need to double escape all the special meta characters as one escape is used for String and second is for regex engine.
This will work:
function validatePhoneNumberFormat(){
var phoneNumber = dijit.byId("phone");
var phoneFormat = /^\d{3}-\d{3}-\d{4}$/;
phoneNumber.validator = function(value){
console.log(value);
console.log(phoneFormat.test(value.trim()));
return phoneFormat.test(value.trim());
}
}
REGEX EXPLANATION
/^\d{3}-\d{3}-\d{4}$/
Assert position at the beginning of the string «^»
Match a single character that is a “digit” (ASCII 0–9 only) «\d{3}»
Exactly 3 times «{3}»
Match the character “-” literally «-»
Match a single character that is a “digit” (ASCII 0–9 only) «\d{3}»
Exactly 3 times «{3}»
Match the character “-” literally «-»
Match a single character that is a “digit” (ASCII 0–9 only) «\d{4}»
Exactly 4 times «{4}»
Assert position at the very end of the string «$»
I'm currently using code like this:
var alphaExp = /^[a-zA-Z]+$/;
if (f_name.value.length > 1 && f_name.value.match(alphaExp)) {
// success
}
else {
document.getElementById("f_name_mark").innerText = "<img src='images/icons/cross.png' class='mark'>";
// fail
}
However, alphaExp only accounts for upper- and lower-case English characters. How can I allow for foreign letters (é, å, ü etc) and spaces/hyphens?
You can try following regex:
/^[a-z- \xC0-\xFF]+$/i
Sample
JSFiddle
function validate() {
var regex = /^[a-z- \xC0-\xFF]+$/i;
var value = document.getElementById("txt").value;
document.getElementById("result").innerHTML = regex.test(value) ? "valid": "incorrect"
}
<input type="text" id="txt">
<button onclick="validate()">Validate</button>
<p id="result"></p>
Encoding Reference
Insert \s to allow spaces.
Add the others letters or symbols you want match.
var alphaexp = /^[a-zA-Z\séåü]+$/
I need to find a reg ex that only allows alphanumeric. So far, everyone I try only works if the string is alphanumeric, meaning contains both a letter and a number. I just want one what would allow either and not require both.
/^[a-z0-9]+$/i
^ Start of string
[a-z0-9] a or b or c or ... z or 0 or 1 or ... 9
+ one or more times (change to * to allow empty string)
$ end of string
/i case-insensitive
Update (supporting universal characters)
if you need to this regexp supports universal character you can find list of unicode characters here.
for example: /^([a-zA-Z0-9\u0600-\u06FF\u0660-\u0669\u06F0-\u06F9 _.-]+)$/
this will support persian.
If you wanted to return a replaced result, then this would work:
var a = 'Test123*** TEST';
var b = a.replace(/[^a-z0-9]/gi, '');
console.log(b);
This would return:
Test123TEST
Note that the gi is necessary because it means global (not just on the first match), and case-insensitive, which is why I have a-z instead of a-zA-Z. And the ^ inside the brackets means "anything not in these brackets".
WARNING: Alphanumeric is great if that's exactly what you want. But if you're using this in an international market on like a person's name or geographical area, then you need to account for unicode characters, which this won't do. For instance, if you have a name like "Âlvarö", it would make it "lvar".
Use the word character class. The following is equivalent to a ^[a-zA-Z0-9_]+$:
^\w+$
Explanation:
^ start of string
\w any word character (A-Z, a-z, 0-9, _).
$ end of string
Use /[^\w]|_/g if you don't want to match the underscore.
/^([a-zA-Z0-9 _-]+)$/
the above regex allows spaces in side a string and restrict special characters.It Only allows
a-z, A-Z, 0-9, Space, Underscore and dash.
^\s*([0-9a-zA-Z]*)\s*$
or, if you want a minimum of one character:
^\s*([0-9a-zA-Z]+)\s*$
Square brackets indicate a set of characters. ^ is start of input. $ is end of input (or newline, depending on your options). \s is whitespace.
The whitespace before and after is optional.
The parentheses are the grouping operator to allow you to extract the information you want.
EDIT: removed my erroneous use of the \w character set.
For multi-language support:
var filtered = 'Hello Привет 你好 123_456'.match(/[\p{L}\p{N}\s]/gu).join('')
console.log(filtered) // --> "Hello Привет 你好 123456"
This matches any letter, number, or space in most languages.
[...] -> Match with conditions
[ab] -> Match 'a' OR 'b'
\p{L} -> Match any letter in any language
\p{N} -> Match any number in any language
\s -> Match a space
/g -> Don't stop after first match
/u -> Support unicode pattern matching
Ref: https://javascript.info/regexp-unicode
This will work
^(?=.*[a-zA-Z])(?=.*[0-9])[a-zA-Z0-9]+$
It accept only alphanumeriuc characters alone:
test cases pased :
dGgs1s23 - valid
12fUgdf - valid,
121232 - invalid,
abchfe - invalid,
abd()* - invalid,
42232^5$ - invalid
or
You can also try this one. this expression satisfied at least one number and one character and no other special characters
^(?=.*[0-9])(?=.*[a-zA-Z])([a-zA-Z0-9]+)$
in angular can test like:
$scope.str = '12fUgdf';
var pattern = new RegExp('^(?=.*[0-9])(?=.*[a-zA-Z])([a-zA-Z0-9]+)$');
$scope.testResult = pattern.test($scope.str);
PLUNKER DEMO
Refered:Regular expression for alphanumeric in Angularjs
Instead of checking for a valid alphanumeric string, you can achieve this indirectly by checking the string for any invalid characters. Do so by checking for anything that matches the complement of the valid alphanumeric string.
/[^a-z\d]/i
Here is an example:
var alphanumeric = "someStringHere";
var myRegEx = /[^a-z\d]/i;
var isValid = !(myRegEx.test(alphanumeric));
Notice the logical not operator at isValid, since I'm testing whether the string is false, not whether it's valid.
I have string similar to Samsung Galaxy A10s 6.2-Inch (2GB,32GB ROM) Android 9.0, (13MP+2MP)+ 8MP Dual SIM 4000mAh 4G LTE Smartphone - Black (BF19)
Below is what i did:
string.replace(/[^a-zA-Z0-9 ,._-]/g, '').split(',').join('-').split(' ').join('-').toLowerCase()
Notice i allowed ,._- then use split() and join() to replace , to - and space to - respectively.
I ended up getting something like this:
samsung-galaxy-a10s-6.2-inch-2gb-32gb-rom-android-9.0-13mp-2mp-8mp-dual-sim-4000mah-4g-lte-smartphone-black-bf19-20 which is what i wanted.
There might be a better solution but this is what i found working fine for me.
Extend the string prototype to use throughout your project
String.prototype.alphaNumeric = function() {
return this.replace(/[^a-z0-9]/gi,'');
}
Usage:
"I don't know what to say?".alphaNumeric();
//Idontknowwhattosay
Even better than Gayan Dissanayake pointed out.
/^[-\w\s]+$/
Now ^[a-zA-Z0-9]+$ can be represented as ^\w+$
You may want to use \s instead of space. Note that \s takes care of whitespace and not only one space character.
Input these code to your SCRATCHPAD and see the action.
var str=String("Blah-Blah1_2,oo0.01&zz%kick").replace(/[^\w-]/ig, '');
JAVASCRIPT to accept only NUMBERS, ALPHABETS and SPECIAL CHARECTERS
document.getElementById("onlynumbers").onkeypress = function (e) {
onlyNumbers(e.key, e)
};
document.getElementById("onlyalpha").onkeypress = function (e) {
onlyAlpha(e.key, e)
};
document.getElementById("speclchar").onkeypress = function (e) {
speclChar(e.key, e)
};
function onlyNumbers(key, e) {
var letters = /^[0-9]/g; //g means global
if (!(key).match(letters)) e.preventDefault();
}
function onlyAlpha(key, e) {
var letters = /^[a-z]/gi; //i means ignorecase
if (!(key).match(letters)) e.preventDefault();
}
function speclChar(key, e) {
var letters = /^[0-9a-z]/gi;
if ((key).match(letters)) e.preventDefault();
}
<html>
<head></head>
<body>
Enter Only Numbers:
<input id="onlynumbers" type="text">
<br><br>
Enter Only Alphabets:
<input id="onlyalpha" type="text" >
<br><br>
Enter other than Alphabets and numbers like special characters:
<input id="speclchar" type="text" >
</body>
</html>
A little bit late, but this worked for me:
/[^a-z A-Z 0-9]+/g
a-z : anything from a to z.
A-Z : anything from A to Z (upper case).
0-9 : any number from 0 to 9.
It will allow anything inside square brackets, so let's say you want to allow any other character, for example, "/" and "#", the regex would be something like this:
/[^a-z A-Z 0-9 / #]+/g
This site will help you to test your regex before coding.
https://regex101.com/
Feel free to modify and add anything you want into the brackets.
Regards :)
It seems like many users have noticed this these regular expressions will almost certainly fail unless we are strictly working in English. But I think there is an easy way forward that would not be so limited.
make a copy of your string in all UPPERCASE
make a second copy in all lowercase
Any characters that match in those strings are definitely not alphabetic in nature.
let copy1 = originalString.toUpperCase();
let copy2 = originalString.toLowerCase();
for(let i=0; i<originalString.length; i++) {
let bIsAlphabetic = (copy1[i] != copy2[i]);
}
Optionally, you can also detect numerics by just looking for digits 0 to 9.
Try this... Replace you field ID with #name...
a-z(a to z),
A-Z(A to Z),
0-9(0 to 9)
jQuery(document).ready(function($){
$('#name').keypress(function (e) {
var regex = new RegExp("^[a-zA-Z0-9\s]+$");
var str = String.fromCharCode(!e.charCode ? e.which : e.charCode);
if (regex.test(str)) {
return true;
}
e.preventDefault();
return false;
});
});
Save this constant
const letters = /^[a-zA-Z0-9]+$/
now, for checking part use .match()
const string = 'Hey there...' // get string from a keyup listner
let id = ''
// iterate through each letters
for (var i = 0; i < string.length; i++) {
if (string[i].match(letters) ) {
id += string[i]
} else {
// In case you want to replace with something else
id += '-'
}
}
return id
Alphanumeric with case sensitive:
if (/^[a-zA-Z0-9]+$/.test("SoS007")) {
alert("match")
}
Also if you were looking for just Alphabetical characters, you can use the following regular expression:
/[^a-zA-Z]/gi
Sample code in typescript:
let samplestring = "!#!&34!# Alphabet !!535!!! is safe"
let regex = new RegExp(/[^a-zA-Z]/gi);
let res = samplestring.replace(regex,'');
console.log(res);
Note: if you are curious about RegEx syntax, visit regexr and either use the cheat-sheet or play with regular expressions.
Edit: alphanumeric --> alphabetical
Only accept numbers and letters (No Space)
function onlyAlphanumeric(str){
str.value=str.value.replace(/\s/g, "");//No Space
str.value=str.value.replace(/[^a-zA-Z0-9 ]/g, "");
}
<div>Only accept numbers and letters </div>
<input type="text" onKeyUp="onlyAlphanumeric(this);" >
Here is the way to check:
/**
* If the string contains only letters and numbers both then return true, otherwise false.
* #param string
* #returns boolean
*/
export const isOnlyAlphaNumeric = (string: string) => {
return /^(?=.*[a-zA-Z])(?=.*[0-9])[a-zA-Z0-9]+$/.test(string);
}
Jquery to accept only NUMBERS, ALPHABETS and SPECIAL CHARECTERS
<html>
<head>
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
</head>
<body>
Enter Only Numbers:
<input type="text" id="onlynumbers">
<br><br>
Enter Only Alphabets:
<input type="text" id="onlyalpha">
<br><br>
Enter other than Alphabets and numbers like special characters:
<input type="text" id="speclchar">
<script>
$('#onlynumbers').keypress(function(e) {
var letters=/^[0-9]/g; //g means global
if(!(e.key).match(letters)) e.preventDefault();
});
$('#onlyalpha').keypress(function(e) {
var letters=/^[a-z]/gi; //i means ignorecase
if(!(e.key).match(letters)) e.preventDefault();
});
$('#speclchar').keypress(function(e) {
var letters=/^[0-9a-z]/gi;
if((e.key).match(letters)) e.preventDefault();
});
</script>
</body>
</html>
**JQUERY to accept only NUMBERS , ALPHABETS and SPECIAL CHARACTERS **
<!DOCTYPE html>
$('#onlynumbers').keypress(function(e) {
var letters=/^[0-9]/g; //g means global
if(!(e.key).match(letters)) e.preventDefault();
});
$('#onlyalpha').keypress(function(e) {
var letters=/^[a-z]/gi; //i means ignorecase
if(!(e.key).match(letters)) e.preventDefault();
});
$('#speclchar').keypress(function(e) {
var letters=/^[0-9a-z]/gi;
if((e.key).match(letters)) e.preventDefault();
});
<html>
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.4.1/jquery.min.js">
Enter Only Numbers:
Enter Only Alphabets:
Enter other than Alphabets and numbers like special characters:
</body>
</html>