How do I allow user to enter only underscore, space, alphabets and numbers in textbox

Question

I have the following jQuery.

    $(document).ready(function() {
$('#name').bind('keypress', function(e) {
    if($('#name').val().length == 0){
        if (e.which == 32){//space bar
            e.preventDefault();
        }
    }
    }); 
});

It accepts any character except space when the length is 0. How do I allow user to enter only a-z A-z 0-9 _ and space(in between characters)?

It's worth noting that if the user was to highlight everything in the textbox, they could press space and that space would be inserted. — George, Sep 04 '13 at 10:48
possible duplicate of [How to prevent user from entering special characters in text box when length is 0?](http://stackoverflow.com/questions/18608954/how-to-prevent-user-from-entering-special-characters-in-text-box-when-length-is) — OneOfOne, Sep 04 '13 at 11:05
I want to restrict user from entering special characters other than underscore(_). So this is not a duplicate. — JavaBeginner, Sep 04 '13 at 11:11

Praveen · Answer 1 · 2013-09-04T12:00:50.320

3

Why don't you try it in regex

/[^A-Za-z0-9_\s]/

Update: Try this

$(document).ready(function () {
    $('#name').bind('keypress', function (e) {
        // Filter non-digits from input value.
        if ($('#name').val().length == 0) {
            if (e.which == 32) { //space bar
                e.preventDefault();
            }
        } else {
            $(this).val($(this).val().replace(/[^A-Za-z0-9_\s]/, ''))
        }
    });
});

Regexplained here

JSFiddle

edited Sep 04 '13 at 12:00

answered Sep 04 '13 at 10:53

Praveen

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let me create a fiddle for better understanding, I'm sure you will enjoy it – Praveen Sep 04 '13 at 11:01

score 1 · Answer 2 · answered Sep 04 '13 at 10:48

function EnableAlphaNumericSpecial(event) {
        // 65 - 90 = smaller case 'a',  32 - space, 48 - 58 - numeric - (0 - 9),  39 - ',35-#, 45- -

        var keyCode = event.charCode;               

        if (keyCode == undefined) 
                keyCode = event.keyCode;

        if ((keyCode >= 65 && keyCode <= 90) || (keyCode >= 97 && keyCode <= 122) || (keyCode >= 48 && keyCode <= 57) || (keyCode == 32) || (keyCode == 39) || (keyCode == 45) || (keyCode == 35) || (keyCode == 0))
                event.returnValue = true;
            else {    //check if it is a IE browser or Firefox browser.
                if (window.event)
                    event.returnValue = false;
                else
                    event.preventDefault();
            }
    }

Something like would help you ..Try this and let me know

@JavaBeginner the code is very similar to what you would have in jQuery. At least try to use it/convert it yourself. After all, jQuery is just javascript convenience methods. — Andrew Grothe, Sep 04 '13 at 11:04

score 0 · Answer 3 · answered Feb 08 '18 at 04:52

Try This...

jQuery(document).ready(function($){
    $('#name').keypress(function (e) {
        var regex = new RegExp("^[0-9a-zA-Z_\. ]+$");
        var str = String.fromCharCode(!e.charCode ? e.which : e.charCode);
        if (regex.test(str)) {
            return true;
        }
        e.preventDefault();
        return false;
    });
});

How do I allow user to enter only underscore, space, alphabets and numbers in textbox

3 Answers3