CODEWITHSUNDEEP
javascriptjqueryinputvalidationalphanumeric
Gereltod
Gereltod

Reputation: 2234

only allow English characters and numbers for text input

Live Demo: http://jsfiddle.net/thisizmonster/DveuB/

How can I change this so that the input only allows the characters A-Z, a-z, 0-9 while typing, without using a regular expression?

Upvotes: 18

Views: 95833

Answers (7)

mohammad asghari
mohammad asghari

Reputation: 1894

All other answers are fine but don't prevent copy and paste non-English characters.

‌Bellow code works for both (key press and copy-paste):

<input type="text" name="text" oninput="this.value=this.value.replace(/[^A-Za-z\s]/g,'');">

Upvotes: 7

vovchisko
vovchisko

Reputation: 2229

I'm using this function.

By modifying RegExp you can get rid of any characters you don't like personally.

$(function(){
    $("#user").bind('keypress',function(e){
        var regex = new RegExp("^[a-zA-Z0-9 ]+$");
        var str = String.fromCharCode(!e.charCode ? e.which : e.charCode);
        if (regex.test(str)) return true;
        e.preventDefault();
        return false;
    });
});

Upvotes: 1

SajithG
SajithG

Reputation: 130

<html>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script>
$(document).ready(function () {

  $('.clsAlphaNoOnly').keypress(function (e) {  // Accept only alpha numerics, no special characters 
        var regex = new RegExp("^[a-zA-Z0-9 ]+$");
        var str = String.fromCharCode(!e.charCode ? e.which : e.charCode);
        if (regex.test(str)) {
            return true;
        }

        e.preventDefault();
        return false;
    }); 
})
</script>

<body>

<input class='clsAlphaNoOnly' type='text'>
</body>
</html>

Upvotes: 2

Siya Matsakarn
Siya Matsakarn

Reputation: 427

<input type="text" id="firstName"  onkeypress="return (event.charCode >= 65 && event.charCode <= 90) || (event.charCode >= 97 && event.charCode <= 122) || (event.charCode >= 48 && event.charCode <= 57)" />

The ASCII Character Set : https://www.w3schools.com/charsets/ref_html_ascii.asp

Upvotes: 10

Bas Slagter
Bas Slagter

Reputation: 9929

Think you have to write some if statements or something like that, that takes the keycode and validates it against some numbers that represent other keycodes:

for example: if(keycode < 80) return false;

Upvotes: 0

321X
321X

Reputation: 3185

You can do something like this: http://jsfiddle.net/DveuB/1/

Then it's only 0-9, a-z and A-Z

$(function(){
    $("#user").keypress(function(event){
        if ((event.charCode >= 48 && event.charCode <= 57) || // 0-9
            (event.charCode >= 65 && event.charCode <= 90) || // A-Z
            (event.charCode >= 97 && event.charCode <= 122))  // a-z
            alert("0-9, a-z or A-Z");
    });
});

Update: http://jsfiddle.net/DveuB/4/
To prevent what @mu is talking about:

$("#user").keyup(function(event){
    if (event.altKey == false && event.ctrlKey == false)
        if ((event.keyCode >= 48 && event.keyCode <= 57 && event.shiftKey== false) ||
            (event.keyCode >= 65 && event.keyCode <= 90) ||
            (event.keyCode >= 97 && event.keyCode <= 122))
            alert("0-9, a-z or A-Z");
});

Upvotes: 2

Paul
Paul

Reputation: 141829

Assuming you also want to accept spaces:

$("#user").keypress(function(event){
    var ew = event.which;
    if(ew == 32)
        return true;
    if(48 <= ew && ew <= 57)
        return true;
    if(65 <= ew && ew <= 90)
        return true;
    if(97 <= ew && ew <= 122)
        return true;
    return false;
});

If you don't want to accept spaces then remove the if(ew == 32) return true;

JSFiddle

Upvotes: 47

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