Decimal is not allowed here. Kindly help

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<script type="text/javascript">
    $(function() {
    $('.textBox_for6tds').keydown(function(e) {
            if (e.shiftKey || e.ctrlKey || e.altKey) {
                e.preventDefault();
            } else {
                var key = e.keyCode;
                if (!((key == 8)  || (key == 46) || (key >= 35 && key <= 40) || (key >= 48 && key <= 57) || (key >= 96 && key <= 105)))
                {
                    e.preventDefault();
                    alert("Please enter numeric value!");

                }
            }
        });
    });

--everything is working fine.
--but if i want to enter 43.53
--the . if i enter its not allowing me to do.......
--what is the ascii of .(decimal). Or how to handle this.

Posted 30-Sep-13 22:11pm

anurag19289

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Comments

Thanks7872 1-Oct-13 5:02am

May i know why you are doing all this? I mean what this code is expected to do?

anurag19289 2-Oct-13 9:38am

Hi rohan-

in the textbox user can enter numbers as well as decimal. but the ascii of decimal was not working. But i got to know that the ascii of '.' is 190 but not 46

4 solutions

Solution 2

You are testing for a strange collection of ASCII values. Your allowed values are:

 8      [backspace]
46      .
35      #
36      $
37      %
38      &
39      '
40      (
48-57   digits 0-9
96      `
97-105  letters a-i

You may want to visit http://www.asciitable.com/[^] to see what you really want.

I'm not sure why you can't type a period, but that seems to be only one of many problems with the code.

Posted 1-Oct-13 2:15am

Brian A Stephens

Updated 1-Oct-13 2:17am

v2

Solution 3

There are JQuery plugins which make this sort of stuff much easier. If you use the filter_input plugin, then you can simply use:

JavaScript

$('.textBox_for6tds').filter_input({ regex: '[0-9.]' });

Posted 1-Oct-13 2:46am

Nick Fisher (Consultant)

v2

Solution 1

see ASCII of decimal "." is 46. and you are checking for 46 that is the reason you are not allowed to enter the decimal.

If you want to allow decimal just remove the condition for 46.
Hope that help you.

-SG

Posted 30-Sep-13 22:28pm

Singh Gyan

Comments

anurag19289 1-Oct-13 4:53am

Hi Gyan-

I have mentioned as not equal to.(!)
if (!((key == 8) || (key == 46) || (key >= 35 && key <= 40) || (key >= 48 && key <= 57) || (key >= 96 && key <= 105)))

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Sampath Kumar Sathiya · Accepted Answer · 2013-10-01T03:44:00

Solution 4

Hi,

The dot in numeric keypad section ASCII value is 46. But the dot in general the key ascii value is 190.

if (!((key == 8) || (key == 46) || (key == 190) || (key >= 35 && key <= 40) || (key >= 48 && key <= 57) || (key >= 96 && key <= 105)))
{
e.preventDefault();
alert("Please enter numeric value!");

}

Posted 1-Oct-13 3:44am

Sampath Kumar Sathiya

Comments

Brian A Stephens 1-Oct-13 10:00am

Now that makes sense!

anurag19289 1-Oct-13 10:12am

This is superb..... +5
NOw this is what i was looking for. Its working perfectly fine now.

anurag19289 6-Nov-13 8:15am

But when i am trying the dot in numlock section[numbers in right side of the keyboard]. Getting the error. But it works fine for the key near comma(,).