function test($arg1, $arg2, $arg3 = "")
{
if(isset($arg3))
{
// do whatever with $arg3
}
}
-----Original Message-----
From: Jonathan Villa [mailto:[EMAIL PROTECTED]
Sent: Thursday, August 21, 2003 7:56 PM
To: [EMAIL PROTECTED]
Subject: [PHP] How to make an argument optional...confirmation...
I want to create a function with an optional argument/parameter but have
never read a concrete answer on how to do it.
This is what I am assuming
function test(arg1,arg2,arg3 = null)
{
arg3 will be optional
}
is this correct?
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