Develop Reference

jQuery prevent writing letters and change comma to dot

I have tried to modify this code but it just wont work...probably some small mistake but i can't debug it :(
// replace , with . and block writing letters
$(document).on("keydown", ".amount", function () {
$(this).keydown(function(e) {
if(e.keyCode==188 || e.keyCode==110 || e.keyCode==108){
e.preventDefault();
$(this).val($(this).val() + '.');
}
var key = e.charCode || e.keyCode || 0;
return (key == 8 || key == 9 || key == 46 || key == 110 || key == 188 || key == 190 || (key >= 35 && key <= 40) || (key >= 48 && key <= 57) || (key >= 96 && key <= 105));
});
};
This is original code and it doesn't work on dynamic content
that is why i want to modify it!

How about just calling your ForceNumericOnly method when the user clicks into an input with .amount?
$(document).on('focus', '.amount', function(){
$(this).ForceNumericOnly();
});
https://jsfiddle.net/daveSalomon/r5n8xuhx/5/
You could (should) optimise it so it doesn't add the keydown handler again if it's already run the ForceNumericOnly code... something like:
$.fn.ForceNumericOnly = function() {
return this.each(function() {
if($(this).data('forceNumberOnly'){ return; }
$(this).data('forceNumberOnly',true);
...
});
};
https://jsfiddle.net/daveSalomon/r5n8xuhx/6/

There's no JS required - just use a number input:
<input class="amount" type="number">

Arrow keys not working in firefox with given javascript

I have one textbox coded like below :
<input name="txttelephone" type="text" maxlength="16" id="txttelephone" class="InputTextBox" onkeypress="return NumbersOnly(event);" required />
And javascript function is like below :
function NumbersOnly(e) {
var unicode = e.charCode ? e.charCode : e.keyCode;
if (unicode != 8) {
if (unicode < 48 || unicode > 57) {
if (unicode == 9)
return true;
else
return false;
}
}
}
Now When I run this in chrome arrow keys working proper but in firefox arrow key is not working. Not getting what is the issue.
Please help me with this.
Thanks,
Dipa

You have to exclude arrow key codes. Try following modification in your code.
function NumbersOnly(e) {
var unicode = e.charCode ? e.charCode : e.keyCode;
if (unicode != 8) {
if (unicode < 48 || unicode > 57) {
if (unicode == 9 || IsArrows(e) )
return true;
else
return false;
}
}
}
function IsArrows (e) {
return (e.keyCode >= 37 && e.keyCode <= 40);
}

JQuery to allow only numeric input in text box using key code

I am trying to allow only numbers [0-9] to be typed in a text box. If an alpha or special character is typed, I do not want it to be shown in the text box. Currently my code is as follows:
$('#TEXTBOX').on("keydown", function(event){
var keyCode = event.which;
if(!((keyCode > 47 && keyCode < 58) || (keyCode > 95 && keyCode < 106) || keyCode == 08)){
event.preventDefault();
}
});
I am having a few problems.
This function is still allows special characters [i.e (SHIFT + 1) gives !, (SHIFT + 2) gives #] I do not want these key combinations to allow insert into text box
I am using magic numbers. I would prefer not to use magic numbers and logic but this is the only way I was able to get the input validation to work.... are there any suggestions on other methods?
My main concern is my first problem with the special characters.

$('#TEXTBOX').on("keydown", function(event){
var keyCode = event.which;
var charCode = (event.charCode) ? event.charCode : ((event.keyCode) ? event.keyCode: ((event.which) ? evt.which : 0));
var char = String.fromCharCode(charCode);
var re = new RegExp("[0-9]", "i");
if (!re.test(char))
{
event.preventDefault();
}
});

Use as
$('#TEXTBOX').on("keydown", function(event){
if(event.shiftKey)
return false;
var keyCode = event.which;
if(!((keyCode > 47 && keyCode < 58) || (keyCode > 95 && keyCode < 106) || keyCode == 08)){
event.preventDefault();
}
});

$(document).ready(function() {
$("#txtboxToFilter").keydown(function (e) {
// Allow: backspace, delete, tab, escape, enter and .
if ($.inArray(e.keyCode, [46, 8, 9, 27, 13, 110, 190]) !== -1 ||
// Allow: Ctrl+A
(e.keyCode == 65 && e.ctrlKey === true) ||
// Allow: home, end, left, right, down, up
(e.keyCode >= 35 && e.keyCode <= 40)) {
// let it happen, don't do anything
return;
}
// Ensure that it is a number and stop the keypress
if ((e.shiftKey || (e.keyCode < 48 || e.keyCode > 57)) && (e.keyCode < 96 || e.keyCode > 105)) {
e.preventDefault();
}
});
});

Worked For me (Only Numbers are allowed)
var key = e.charCode || e.keyCode || 0;
alert(key);
if (key < 48 || key > 58)
return false;

Why my condition is not satisfied?

I have a textbox and i want to allow users only enter 4 digits as i want to take time from the user, but i am facing the strange problem in condition.
Fiddle Demo
Javascript
function CheckLength(val, key) {
var keycode = (key.which) ? key.which : key.keyCode;
if(!(keycode == 8 || keycode == 46) && (keycode < 48 || keycode > 57))
return false;
if (val.length < 4)
console.log(val);
else
return false;
}
HTML Markup
<input type="text" id="timepick" onkeyup="return CheckLength(this.value,event);" />
Can anyone help me? why this is happening?
Thanks for your precious time.

You could listen for a "keydown" event instead of "keyup" Heres your Fiddle, (i only changed up to down). , or remove the last entered key by resetting the value in the "keyup" event.
document.getElementById("timepick").addEventListener("keyup", function (e) {
var keycode = (e.which) ? e.which : e.keyCode;
if (e.target.value.length <= 4) console.log(e.target.value)
if (!(keycode == 8 || keycode == 46) && (keycode < 48 || keycode > 57) || e.target.value.length > 3)
e.target.value = e.target.value.substr ( 0,4)
});
Like in this Fiddle
Or you could use 2 seperate events, one "keydown" to prevent the input, and a keyup to read the value
document.getElementById("timepick").addEventListener("keydown", function (e) {
var keycode = (e.which) ? e.which : e.keyCode;
if (!(keycode == 8 || keycode == 46) && ((keycode < 48 || keycode > 57) || e.target.value.length > 3) e.preventDefault()
});
document.getElementById("timepick").addEventListener("keyup", function (e) {
if (e.target.value.length == 4) console.log(e.target.value)
});
Like in this Fiddle
Update, regarding your comment
You could, of course use only a "keydown" event, and build the value you want on your own.
document.getElementById("timepick").addEventListener("keydown", function (e) {
var keycode = (e.which) ? e.which : e.keyCode;
if (!(keycode == 8 || keycode == 46) && ((keycode < 48 || keycode > 57) || e.target.value.length > 3)) e.preventDefault()
if (e.target.value.length == 3) console.log(e.target.value + String.fromCharCode(keycode))
});
Like in this Fiddle

Hope this helps you. It prevents more than 4 values and shows the 4 entered into the log (in the fiddle I change the console.log by an alert).
I solved it by storing the 4 values into a variable and if the user enters more than 4 values restore the textbox value with the variable value:
http://fiddle.jshell.net/6czXu/7/
var vals;
function CheckLength(val, key) {
var keycode = (key.which) ? key.which : key.keyCode;
if (val.length < 5){
alert(val);
// Here store the 4 values
vals = val;
if((keycode == 8 || keycode == 46) && (keycode < 48 || keycode > 57))
return false;
}
else{
// Here we have more than 4 digits entered so
// we restore the prevously stored into vals
document.getElementById("timepick").value = vals;
return false;
}
}

Use the onkeypress event instead and everything should work fine. onkeyup is to late.
<input type="text" id="timepick" onkeypress="return CheckLength(this.value,event);" />

jquery - Numeric Only Text Field

I found this very short clean code to only allow numeric chars in a text field. Currently it only covers numbers 0-9 and backspace and delete. I wanted it to also include decimal/period, so I have been fighting with this to simply include keycode 110 and/or 190. I can not get it to work. Can anyone see what I am doing wrong?
$(document).ready(function() {
$('input.numberinput').bind('keypress', function(e) {
return ( e.which!=8 && e.which!=0 && (e.which<48 || e.which>57) ) || (e.which!=110) ? false : true ;
});
});
jsfiddle here: http://jsfiddle.net/justmelat/EN8pT/
html
<div class="label">Enter a number:</div>
<input type="text" name="txtNumber1" id="txtNumber1" value="" class="numberinput" />
<div class="label">Enter a number:</div>
<input type="text" name="txtNumber2" id="txtNumber2" value="" class="numberinput" />
</div>

Try:
$(document).ready(function () {
$('input.numberinput').bind('keypress', function (e) {
return !(e.which != 8 && e.which != 0 &&
(e.which < 48 || e.which > 57) && e.which != 46);
});
});​
JsFiddle: http://jsfiddle.net/EN8pT/1/

With the above answer you can still do 0.23.12.33 which is not a valid number.
http://www.texotela.co.uk/code/jquery/numeric/ is a great little lightweight plugin that I have used a lot. It takes the pain out of the above.

$("#input").keydown(function(event) {
var theEvent = event || window.event;
var key = theEvent.keyCode || theEvent.which;
// Allow: backspace, delete, tab, escape, and enter
if ( key == 46 || key == 8 || key == 9 || key == 27 || key == 13 || key == 110 || key == 190 ||
// Allow: Ctrl+A
(key == 65 && theEvent.ctrlKey === true) ||
// Allow: home, end, left, right
(key >= 35 && key <= 39)) {
// let it happen, don't do anything
return;
}
else {
// Ensure that it is a number and stop the keypress
if (theEvent.shiftKey || (key < 48 || key > 57) && (key < 96 || key > 105 )) {
theEvent.preventDefault();
}
}
});
with keycode 110 and 190