Checking whether a string is all digits

[Christopher Benson-Manica]

I'm trying to check whether a string is all digits. This part is
easy:

function allDigits( str ) {
var foo=str.split( '' ); // better than charAt()?
for( var idx=0; idx < foo.length; idx++ ) {
if( !isDigit(foo[idx]) ) {
return false;
}
}
return true;
}

I'm not sure about how to implement isDigit(). Is this the best way?

function isDigit( s ) {
if( s.length > 1 ) {
return false;
}
var nums='123456789 0';
return nums.indexOf(s) != -1;
}

--
Christopher Benson-Manica | I *should* know what I'm talking about - if I
ataru(at)cybers pace.org | don't, I need to know. Flames welcome.

[cwdjrxyz]

Christopher Benson-Manica wrote:

I'm trying to check whether a string is all digits.

There likely are several ways to do it. I made a test page for you at
my site at http://www.cwdjr.info/test/detectNonNumerical.html . I have
used this method on a perpetual calendar page of mine, and it has
worked on several of the most recent browsers. I detect if the number
is too large and too small as well as check for a negative number. It
is easy to dump these extras if you are interested only in digits.

[Stephen Chalmers]

Vic Sowers <Mail@Vic_NOSPA M_Sowers.com> wrote in message
news:42******** *************** @news.hal-pc.org...

"Christophe r Benson-Manica" <at***@nospam.c yberspace.org> wrote in message news:d8******** **@chessie.cirr .com...

var foo=str.split( '' ); // better than charAt()?

Actally it isn't. charAt() gives access to individual characters in an
existing string, so why build a new array to do the same thing? Rather
than searching for the character in a string of all characters, check
that its value is within range (which is what the RegExp engine will
do, only much faster).

function allDigits( str )
{
var justDigits=fals e;

if(typeof str!='undefined ' && str.length)
{
for( var i=0, d ; i<str.length && (d=str.charAt(i ))>='0' &&
d<='9' ; i++ )
;
if(i==str.lengt h)
justDigits=true ;
}
return justDigits;
}
--
Stephen Chalmers
547265617375726 520627572696564 206174204F2E532 E207265663A2054 5132343739
3134

[Michael Winter]

On 07/06/2005 08:07, RobG wrote:

[snip]

Try this:

alert( "".split(/\d/).length );
alert( "1".split(/\d/).length );

Firefox reports: 1 then 2, but IE reports 1 then 0.
As usual, IE is broken.

The character class escape, \d, cannot match an empty string, so the
return value should be an array that contains one element: the empty
string. If a pattern can match the empty string (for example, \d|), then
the result should be an array with zero elements.

The character class escape, \d, can match the digit, 1. Therefore the
result, in this particular case, should be an array that contains two
elements. The first is all of the characters from the start of the
string to just before the matched digit (an empty string), and all of
the characters after the matched digit to the end of the string (again,
an empty string).

If the second regular expression captured the digit, (\d), the resulting
array would have three elements: '', '1', ''.
Explain that and keep a straight face.
Not a straight face. More one of resigned expectation.

[snip]
y = [ ,,, ]; alert(y.length) // Firefox says 3, IE says 4
y = [ ,,,'' ]; alert(y.length) // Firefox says 4, IE says 4

Again, IE is wrong. Each elision (just a comma; no value) increments the
length of the array. In the first literal, there are three elisions so
the length is three. In the second literal there are also three
elisions, however the string literal is appended to the array[1],
creating the fourth element.

[snip]

Mike
[1] That's not how the specification describes it, but it's the same
result.

--
Michael Winter
Replace ".invalid" with ".uk" to reply by e-mail.

[Lasse Reichstein Nielsen]

RobG <rg***@iinet.ne t.auau> writes:

You think that's weird? Try this:

alert( "".split(/\d/).length );
alert( "1".split(/\d/).length );

Firefox reports: 1 then 2, but IE reports 1 then 0. Explain that and
keep a straight face.
IE is wrong. Hard not to say that with a straight face. :)

It seems IE removes initial/terminal empty parts, if it does any split
at all.

I was wrong too. It is possible to get a zero-length result out of split:
"".split(/.?/);
(the empty string with a pattern that matches a zero-length string).
Here's another:

y = [ ]; alert(y.length) // Both browsers say 0
y = [ ,,, ]; alert(y.length) // Firefox says 3, IE says 4
Again IE fails to satisfy the ECMAScript standard. Firefox is correct.
If the last element is undefined, Firefox doesn't add it to the array.

No, it's not that there is an element that is undefined. The "," sequence
is an "elision" in the grammar of array literals, and a different token
than the one for actual elements.

/L
--
Lasse Reichstein Nielsen - lr*@hotpop.com
DHTML Death Colors: <URL:http://www.infimum.dk/HTML/rasterTriangleD OM.html>
'Faith without judgement merely degrades the spirit divine.'

[Dr John Stockton]

JRS: In article <mJ1pe.891$2H2. 647@trndny08>, dated Mon, 6 Jun 2005
19:45:22, seen in news:comp.lang. javascript, drWot <dr***@yankeewe b.com>
posted :

String.prototy pe.isdigits=fun ction(){
return (/\D/.test(this)==fa lse);
}

There is no need to use ==false; return !/\D/.test(this) .

That accepts an empty string.

--
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