I am still new to javascript and I am trying to validate my form.
One of my inputs is a text input for an identity number that follows the following pattern: ####XX where # represents a number and X represents a capital letter from A-Z.
Here is my code so far:
var IDnum = document.getElementById('identityNumber').value;
if ( (isNaN(IDnum.charAt(0))) && (isNaN(IDnum.charAt(1)))&& (isNaN(IDnum.charAt(2))) && (isNaN(IDnum.charAt(3))) && (!isNaN(IDnum.charAt(4))) )
{
document.getElementById('identityError').style.display = "inline-block";
}
else
{
document.getElementById('identityError').style.display = "none";
}
I have tried to google it and have seen some info where they use a RegExp however i have yet to learn anything like that.
With my code above, no matter what i type it, it still validates it. Any ideas what i am doing wrong and if there is a more simple and easier way?
EDIT: after looking to regex and similar answers the following
^\d{4}[A-Z]{2}$
did not work either
A regular expression is the way to go here. Use the pattern ^\d{4}[A-Z]$:
document.querySelector('button').addEventListener('click', (e) => {
const { value } = document.querySelector('input');
if (value.match(/^\d{4}[A-Z]$/)) {
console.log('OK');
} else {
console.log('Bad');
}
});
<input>
<button>submit</button>
^\d{4}[A-Z]$ means:
^ - Match the start of the string
\d{4} - Match a digit character (0 to 9) 4 times
[A-Z] - Match a character from A to Z
$ - Match the end of the string
You can use regular expression to identify whether string has 4 digits before a character.
each \d represents a digit, \d\d\d\d means 4 digits (alternatively \d{4}).
followed by . means 4 digits followed by any character.
function isAllowed(str) {
return str.match(/^\d\d\d\d.$/g) !== null
}
console.log(isAllowed("1234X"));
console.log(isAllowed("123a"));
console.log(isAllowed("3892#"));
console.log(isAllowed("X"));
var IDnum = document.getElementById('identityNumber').value;
if (isAllowed(IDnum))
{
document.getElementById('identityError').style.display = "inline-block";
}
else
{
document.getElementById('identityError').style.display = "none";
}
function RegexCheck(str) {
var pettern = new RegExp('^[0-9]{4,}[A-Z]{1,}');
return pettern.test(str);
}
console.log(RegexCheck("####X"));
console.log(RegexCheck("1234A"));
console.log(RegexCheck("2C35B"));
console.log(RegexCheck("A698C"));
console.log(RegexCheck("1698b"));
You can use the pattern attribute to provide a RegExp string:
^\d{4}[A-Z]{2}$ would be a string consisting of 4 digits followed by two capital letters between A and Z.
Explanation
^: Beginning of the string.
\d{4}: Exactly 4 digits in a row (this could also be written as \d\d\d\d)
[A-Z]{2}: Exactly 2 characters from the range of character between A and Z (alternatively [A-Z][A-Z]).
$: The end of the string.
input:invalid {
color: red;
}
input:not(:invalid) {
color: green;
}
<input type="text" pattern="^\d{4}[A-Z]{2}$">
I'm trying to check if there's a number in a text input using regular expression. Here's the code:
var regex = /^[0-9]+$/;
if (myInput.val().match(regex)) {
console.log("number");
} else {
console.log("bad");
}
It works well, but if I add text, then backspace all the way, I get "bad". How can I make it log "good" when there isn't anything in the text input? I don't want to allow spaces, but I want to allow an empty input.
I tried:
var regex = /\s ^[0-9]+$/;
But then whatever I insert in the input, I always get "bad".
This might fit , either you test for your Exp (^[a-zA-Z0-9]+$) or for an empty string (^$).
var regex = /(^$)|(^[a-zA-Z0-9]+$)/;
if (myInput.val().match(regex)) {
console.log("number");
} else {
console.log("bad");
}
try this (* in place of +)
var regex = /^[0-9]*$/;
if (myInput.val().test(regex)) {
console.log("number");
} else {
console.log("bad");
}
I want to validate a text field (first name) using javascript. such that it should only contain text. NO special characters and No numbers.and since it is just the first name it should only contain one word. (no spaces)
Allowed:
John
john
Not Allowed
john kennedy.
John kennedy.
john123.
123john.
I tried this but its not working.
if( !validateName($fname))
{
alert("name invalid");
}
function validateName($name) {
var nameReg = /^A-Za-z*/;
if( !nameReg.test( $name ) ) {
return false;
} else {
return true;
}
}
EDIT:
I tried
var nameReg = /^[A-Za-z]*/;
but it still doesn't show the alert box when I enter john123 or 123john.
nameReg needs to be /^[a-z]+$/i (or some varient). The ^ matches the start of the string and $ matches the end. This is "one or more a-z characters from the start to the end of the string, case-insensitive." You can change + to *, but then the string could be empty.
http://jsfiddle.net/ExplosionPIlls/pwYV3/1/
Use a character class:
var nameReg = /^[A-Za-z]*/;
Without the containing [] (making it a character class), you're specifying a literal A-Za-z.
UPDATE:
Add a $ to the end of the Regex.
var nameReg = /^[A-Za-z]*$/;
Otherwise, john123 returns valid as the Regex is matching john and can ignore the 123 portion of the string.
Working demo: http://jsfiddle.net/GNVck/
Trying to get the correct regex for this - only letters, spaces, hypens, and commas. So far this only works if you only input 1 charactor. Any more then that, and it returns false. Anyone able to help?
$('#submit').click(function () {
var locationtest = /[^a-zA-Z \-\.\,]/;
if (!locationtest.test($('#location').val())) {
alert('Nope, try again!');
$('#location').val('')
return false;
} else {
alert('You got it!');
}
});`
This should do it, it matches 1 or more characters within the set you described
/^[a-zA-Z \-\,]+$/
I took out the \., your description says letters, spaces, hyphens, commas
You're close, you just need to specify how many times you want the character to appear.
The following code would specify 0 or more times
var locationtest = /[^a-zA-Z -.\,]*/;
And this code would specify 1 or more times
var locationtest = /[^a-zA-Z -.\,]+/;
The importance being the * and + characters.
Add a quantifier + and the global flag /g:
var locationtest = /[^a-zA-Z \-\.\,]+/g;
Your expression is correct, you just need to invert the match result.
/[^a-zA-Z \-\.\,]/
Will match if the string contains any char that is not what you want (the leading ^ in the character class).
I.e remove the !:
var locationtest = /[^a-zA-Z \-\.\,]/;
if (locationtest.test($('#location').val())) {
alert('Nope, try again!');
$('#location').val('')
return false;
} else {
alert('You got it!');
}
Note that empty string will pass as valid, if you don't want that, you can use this instead:
/[^a-zA-Z \-\.\,]|^$/
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>