RE: [PHP] An integer?

Joao Prado Maia writes: > At 08:18 PM 8/21/2000 +0100, you wrote: > > > > This may be a silly question, but I can't spot and easy way to do it, I > > > want to check if the value inside a vriable is an integer. If I do > > > is_int($variable) I always get false, because this variable qwas read in > > > from a form so its a string, if I make the variable an integer using > > > settype() I just end up with 0 if it was invalid, the number otherwise. I > > > need something that will actually tell me if there the conversion was > > > successfull. Am I missing something? > > > >if(!((int)$variable)) { > > //Not an int or 0 > >} else { > > // An int != 0 > >} > > He can also use gettype() > > if (gettype($variable) == "integer") { > // do stuff > } else { > // do other stuff > } > > http://www.php.net/manual/function.gettype.php > > Regards, > Joao No, neither of these examples will do what he wanted--he's trying to find out whether a given variable (a string, since it came from a form) represents an integer. The gettype() test above will always evaluate to false, since form data is always passed as a string (or an array of strings). The 'if(!((int)$variable))' just tests whether the result of converting the incoming string to an int is a true value or not. Among other things, this means that it doesn't consider '0' an integer, but does consider '45otherstuff' and integer (since the conversion of '45otherstuff' to int results in '45'. -- +----------------------------------------------------------------+ |Torben Wilson <torben@php.net> Netmill iTech| |http://www.coastnet.com/~torben http://www.netmill.fi| |Ph: 1 250 383-9735 torben@netmill.fi| +----------------------------------------------------------------+