Numeric field validation fails for single character

I am using following function to check for my input
my input is like 0.00 i dont want to allow any specialcharacters.

function numericValidation() {
var x = this;
var reg = new RegExp('^[0-9]*\.?[0-9]*$');

if (reg.test(x)) {
return true;
}
else return false;
}

here problem is when i type one character other than this range Regular
expression doesnt work i mean it gives true for input '@' if i type
'@@'then it gives false

actually it shd gv false for both the case.

nkle004 (AT) gmail (DOT) com wrote:

I am using following function to check for my input
my input is like 0.00 i dont want to allow any specialcharacters.

function numericValidation() {
var x = this;
var reg = new RegExp('^[0-9]*\.?[0-9]*$');
---------------------------------------------^

When calling RegExp as a constructor, quoted special characters in the
string literal need two backslashes - you have to quote the quote to get
it past the parser (if that makes sense...). Use:

var reg = new RegExp('^[0-9]*\\.?[0-9]*$');


Note use of '\\'.


[...]


--
Rob

"RobG" <rgqld (AT) iinet (DOT) net.au> wrote

nkle004 (AT) gmail (DOT) com wrote:
I am using following function to check for my input
my input is like 0.00 i dont want to allow any specialcharacters.

function numericValidation() {
var x = this;
var reg = new RegExp('^[0-9]*\.?[0-9]*$');
---------------------------------------------^

When calling RegExp as a constructor, quoted special characters in the
string literal need two backslashes - you have to quote the quote to get
it past the parser (if that makes sense...). Use:

var reg = new RegExp('^[0-9]*\\.?[0-9]*$');


Note use of '\\'.

Or use a regular expression literal:
var reg = /^[0-9]*\.?[0-9]*$/;

[...]


--
Rob

if i use a regular expression literal can i still use test methos with
it ?:

var reg = /^[0-9]*\.?[0-9]*$/;

Vic Sowers wrote:

"RobG" <rgqld (AT) iinet (DOT) net.au> wrote in message
news:0cRNf.599$Nv4.93706 (AT) news (DOT) optus.net.au...
nkle004 (AT) gmail (DOT) com wrote:
I am using following function to check for my input
my input is like 0.00 i dont want to allow any specialcharacters.

function numericValidation() {
var x = this;
var reg = new RegExp('^[0-9]*\.?[0-9]*$');
---------------------------------------------^

When calling RegExp as a constructor, quoted special characters in the
string literal need two backslashes - you have to quote the quote to get
it past the parser (if that makes sense...). Use:

var reg = new RegExp('^[0-9]*\\.?[0-9]*$');


Note use of '\\'.


Or use a regular expression literal:
var reg = /^[0-9]*\.?[0-9]*$/;


[...]


--
Rob

nkle004 (AT) gmail (DOT) com wrote:

if i use a regular expression literal can i still use test methos with
it ?:
var reg = /^[0-9]*\.?[0-9]*$/;

Yes, you can. The RegExp literal creates a RegExp object. Why do you not
just try it?

You can even provide attribution of quoted material, and stop top-posting.

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