Best way to restrict a text field to numbers only?

Question

I'm using the following Javascript to restrict a text field on my website to only accept numerical input, and no other letters or characters. The problem is, it REALLY rejects all other key inputs, like ctrl-A to select the text, or even any other browser functions like ctrl-T or ctrl-W while the text box is selected. Does anyone know of a better script to only allow numerical input, but not block normal commands (that aren't being directly input into the field)? Thanks Here is the code I'm using now:

function numbersonly(e, decimal) 
{
    var key;
    var keychar;

    if (window.event) 
        key = window.event.keyCode;
    else if (e) 
        key = e.which;
    else 
        return true;

    keychar = String.fromCharCode(key);

    if ((key==null) || (key==0) || (key==8) ||  (key==9) || (key==13) || (key==27))
       return true;     
    else if ((("0123456789").indexOf(keychar) > -1))
       return true;
    else if (decimal && (keychar == "."))
       return true;        
    else
       return false;
}

Edit: None of the solutions provided have solved my problem of allowing commands like ctrl-A while the text box is selected. That was the whole point of my asking here, so I have gone back to using my original script. Oh well.

Answer 1

I use this:

    oEl.keypress(function(ev)
    {
        var sKey = String.fromCharCode(ev.which);
        if (!sKey.match(/[0-9]/) || !sKey === "") 
            ev.preventDefault();            
    });

The advantage is, that every key which does not provide an input to the field is still allowed, so you don't have to worry about every single special key. Even combos like CTRL + R do still work.

EDIT As this is not working in Firefox I had to modify the function a little:

    oEl.keypress(function(ev)
    {
        var iKeyCode = ev.which || ev.keyCode;
        var aSpecialKeysForFirefox = [8, 9, 13, 27, 37, 38, 39, 40, 46];
        var sKey = String.fromCharCode(iKeyCode);
        if (sKey !== "" && $.inArray(iKeyCode, aSpecialKeysForFirefox ) < 0 && !sKey.match(/[0-9]/)) {
            ev.preventDefault();
        }
    });

Explanation All Browsers handle jquerys keypress event differently. To make it work in FF the $.inArray check is added. As firefoxs keypress-event doesn't trigger when combinations like strg+tab are used, but the others do, the key.match approach still adds a little value to the latter, as it enables those combinations.

Answer 2

This is a variation on Robert's answer that allows a single decimal point to be entered. If a decimal point has already been entered, only numbers are accepted as input.

JSFiddle - decimal number input

// Allow only decimal number input

$('#decimalInput').keypress(function (e) {
    var a = [];
    var k = e.which;

    for (i = 48; i < 58; i++)
    a.push(i);

    // allow a max of 1 decimal point to be entered
    if (this.value.indexOf(".") === -1) {
        a.push(46);
    }

    if (!(a.indexOf(k) >= 0)) e.preventDefault();

    $('span').text('KeyCode: ' + k);
});

Answer 3

It's worth pointing out that no matter how tightly you manage to control this via the front end (Javascript, HTML, etc), you still need to validate it at the server, because there's nothing to stop a user from turning off javascript, or even deliberately posting junk to your form to try to hack you.

My advice: Use the HTML5 markup so that browsers which support it will use it. Also use the JQuery option previously suggested (the inital solution may have flaws, but it seems like the comments have been working through that). And then do server-side validation as well.

Answer 4

You can handle te event on html by introducing keypresshandler function

function keypresshandler(event)
{
     var charCode = event.keyCode;

     //You condition
     if (charCode == 58 ){

        event.preventDefault();

        return false;
     }        
} 

Answer 5

Here is my solution: a combination of the working ones below.

var checkInput = function(e) {
        if (!e) {
            e = window.event;
        }

        var code = e.keyCode || e.which;

        if (!e.ctrlKey) {

            //46, 8, 9, 27, 13 = backspace, delete, tab, escape, and enter
            if (code == 8 || code == 13 || code == 9 || code == 27 || code == 46)
                return true;
            //35..39 - home, end, left, right
            if (code >= 35 && code <= 39)
                return true;
            //numpad numbers
            if (code >= 96 && code <= 105)
                return true;
            //keyboard numbers
            if (isNaN(parseInt(String.fromCharCode(code), 10))) {
                e.preventDefault();
                return false;
            }
        }
        return true;
    };

Answer 6

This works in IE, Chrome AND Firefox:

<input type="text" onkeypress="return event.charCode === 0 || /\d/.test(String.fromCharCode(event.charCode));" />