php传递一个并不总是在用户函数中设置的变量

Question

i have created a simple function in php, but i'm not sure about how to pass a (optional) variabile if this is not always set, for example:

myfunction( $var1, $var2, $_POST[example] );

as you can see the myfunction is always called but the $_POST[example] isset/exist only when the form was already sent.

my question is, is that correct??? if $_POST[example] don't exist i see no error/notice (Notice: Undefined variable..) but i'm not sure...

maybe is better like this?

 myfunction( $var1, $var2, isset($_POST[example]) );

Answer 1

As AlmaDoMundo's answer shows: make the argument optional:

function myFunction($arg1, $arg2, $arg3 = null)
{
}

Then, to avoid warnings (undefined index), you'll have to call the function like so:

if (isset($_POST['example']))
{
    myFunction($var1, $var2, $_POST['example']);
}
else
{
    myFunction($var1, $var2);
}

If, for some reason you're unable to change the function definition (and therefore, it's signature), just call the function like this:

$var3 = isset($_POST['example']) ? $_POST['example'] : null;
myFunction($var1, $var2, $var3);//null is a value, so you're passing 3 arguments

Calling the your function the second way myFunction($var1, $var2, isset($_POST['example'])); is not the way forward. Just look at isset's signature

bool isset ( mixed $var [, mixed $... ] )

it returns a bool (true/false), not the value of the variable itself. If the key exists, your function will receive true as a third argument, if not false. You want to pass the actual value, so you'll have to either wrap your code in if/else branches, or use a ternary (myFunction($var1, $var2, isset($_POST['example']) ? $_POST['example'] : null);.
Personally, I'd not use a ternary here, because the first example of how to call the function I gave here is just much more readable (it's important to write maintainable code!)

On the notice:
You say you don't see any errors/warnings if you pass $_POST['example'] (you should quote the 'example', BTW, because that should issue a warning, too). Try changing your ini settings to:

error_reporting  =  E_ALL | E_STRICT
#and
display_errors = 1
display_startup_errors = 1 # careful here!

Then, you'll see how many issues your code really has. Check the manual to see what each setting does, how and when to use them.