Ahhh, in that context, with is_integer(), i think
you're partially right. however, i looked it up, and
is_numeric() will work fine.
http://www.php.net/manual/en/function.is-numeric.php
btw-- as for the regexps, none for me, thanks ;)
those things give me the heebie-jeebies.
--- "scott [gts]" <[EMAIL PROTECTED]> wrote:
> i think he's asking for a numeric test, not an
> integer test.
>
> your example looks like it should fail "3.50" (for
> example),
> even though "3.50" is a perfectly valid numeric
> value.
>
> personally, i think you should just use regexps. ;-)
>
> i tried to write a little is_num() type function,
> but i kept
> getting caught on split(); it seems that split()
> converts
> everything to string, no matter what. so if you try
> to split
> on the decimal point of 45.67 ... you get two
> strings, "45" and "67"
> not 45 and 67, so subsequent is_integer() calls fail
>
> (since is_integer() checks the *type* of the
> variable,
> not the *value* -- very important to keep in mind)
>
> here's what i came up with... perhaps it will be of
> some
> help to you (even though it doesnt work)
>
>
> function is_num($number) {
>
> // $x contains number, $y contains any decimal
> value
> list ($x, $y) = split("\.", $number);
>
> // if there's no decimal value
> if ( empty($y) ) {
> return is_int($x);
> }
> else {
> return ( is_int($x) && is_int($y) );
> }
>
> }
>
>
> > -----Original Message-----
> > From: daniel james [mailto:[EMAIL PROTECTED]]
> > Subject: Re: [PHP] is_numeric for php3
> >
> > do you mean, as in,
> >
> > if !is_integer($var) {
> > print("$var is not an integer");
> > }
> >
>
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