JQuery allow only two numbers after decimal point

user676567 picture user676567 · Feb 15, 2014 · Viewed 76.3k times · Source

I found the following JQuery function here which prevents a user from entering anything that's not a number or a single decimal. The function works well but I'd like to improve it to prevent the user from entering 3 or more decimal places i.e. disallow 99.999 and allow 99.99. Any ideas?

 function checkForInvalidCharacters(event, inputBox){                            
     if ((event.which != 46 || inputBox.val().indexOf('.') != -1) && (event.which < 48 || event.which > 57)) {          
                event.preventDefault();           
            }   
     };

Answer

laaposto picture laaposto · Feb 15, 2014

The logic is every time a user entering a number you have to check two things.

  1. Has the user entered decimal point?
  2. Are the decimal places more than two?

For the first one you can use $(this).val().indexOf('.') != -1

For the second one you can use $(this).val().substring($(this).val().indexOf('.'), $(this).val().indexOf('.').length).length > 2

EDIT-1
Also, you have to add event.which != 0 && event.which != 8 so that arrow keys and backspace work in Firefox (Manoj comment)

EDIT-2
Also, you have to add $(this)[0].selectionStart >= text.length - 2 so that you can add digits if the cursor is to the left of the decimal point (BIdesi comment)

EDIT-3
Also, you have to check if user deleted . and placed it somewhere else creating a value with more than 2 digits after the decimal. So you have to add $this.val($this.val().substring(0, $this.val().indexOf('.') + 3)); for cutting extra digits (Gilberto Sánchez comment)

EDIT-4
To handle pasted data, you must bind a paste event handler.Then you have to check if pasted data have . withtext.indexOf('.') > -1 and more than 2 digits after the decimal with text.substring(text.indexOf('.')).length > 3. If so, you have to cut extra digits. Also you have to check that user entered numeric input with $.isNumeric() (darasd comment).

Here is the code:

$('.number').keypress(function(event) {
    var $this = $(this);
    if ((event.which != 46 || $this.val().indexOf('.') != -1) &&
       ((event.which < 48 || event.which > 57) &&
       (event.which != 0 && event.which != 8))) {
           event.preventDefault();
    }

    var text = $(this).val();
    if ((event.which == 46) && (text.indexOf('.') == -1)) {
        setTimeout(function() {
            if ($this.val().substring($this.val().indexOf('.')).length > 3) {
                $this.val($this.val().substring(0, $this.val().indexOf('.') + 3));
            }
        }, 1);
    }

    if ((text.indexOf('.') != -1) &&
        (text.substring(text.indexOf('.')).length > 2) &&
        (event.which != 0 && event.which != 8) &&
        ($(this)[0].selectionStart >= text.length - 2)) {
            event.preventDefault();
    }      
});

$('.number').bind("paste", function(e) {
var text = e.originalEvent.clipboardData.getData('Text');
if ($.isNumeric(text)) {
    if ((text.substring(text.indexOf('.')).length > 3) && (text.indexOf('.') > -1)) {
        e.preventDefault();
        $(this).val(text.substring(0, text.indexOf('.') + 3));
   }
}
else {
        e.preventDefault();
     }
});
.number {
  padding: 5px 10px;
  font-size: 16px;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input type="text" class="number" />