Numeric field validation fails for single character

[nkle004]

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.

[RobG]

nk*****@gmail.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

[Vic Sowers]

"RobG" <rg***@iinet.net.au> wrote in message
news:0c*****************@news.optus.net.au...

nk*****@gmail.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]

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" <rg***@iinet.net.au> wrote in message
news:0c*****************@news.optus.net.au...

nk*****@gmail.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

[Thomas 'PointedEars' Lahn]

nk*****@gmail.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.

<URL:http://safalra.com/special/googlegroupsreply/>
PointedEars