function foo ($arg1_required, $arg2_options = ) {
if (empty($arg2_optional) {
print not passed;
} else {
print explicit;
}
}
Or you can go the other route.
Jordan
Rodent Of Unusual Size [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
An odd request.. I want to have optional function arguments
that *aren't* defined unless values are explicitly passed. Sorta
the equivalent of
function foo($arg1_required, $arg2_optional=undef)
if (isset($arg2_optional)) {
print explicit;
}
else {
print not passed;
]
Subject: [PHP] 'undef' as an argument value
An odd request.. I want to have optional function arguments
that *aren't* defined unless values are explicitly passed. Sorta
the equivalent of
function foo($arg1_required, $arg2_optional=undef)
if (isset($arg2_optional)) {
print explicit
Rick Emery wrote:
function foo($arg1_required, $arg2_optional=)
if ($arg2_optional==) {
print not passed;
}
else {
print explicit;
}
}
No, I said 'undef' because I mean 'undefined'. The argument
can have *any* value; I need to know when it has *none*.
http://www.php.net/manual/en/function.func-num-args.php will return the
number of arguments passed to the function.
Kirk
-Original Message-
From: Rodent of Unusual Size [mailto:[EMAIL PROTECTED]]
Sent: Monday, March 11, 2002 2:25 PM
To: [EMAIL PROTECTED]
Subject: Re: [PHP] 'undef
Johnson, Kirk wrote:
http://www.php.net/manual/en/function.func-num-args.php will
return the number of arguments passed to the function.
Bingo! That will do the trick.. Thanks *very* much!
--
#kenP-)}
Ken Coar, Sanagendamgagwedweinini http://Golux.Com/coar/
Author, developer,
6 matches
Mail list logo