Hello
If I defined a function with 4 parameters but only pass 3 I get an
error. Is there anyway around this?
I want to be able to set the missing parameter to a default value if it
is not passed which works ok but How do I get rid of the error message?
Thanks
Giles Roadnight
http://giles.roadnight.name
If you want an argument to be optional and still be able to check whether or not it's been set, you can use NULL as the default value:
function foobar ($a, $b, $c = null)
{
if (isset($c)) {
echo 'The third argument was set';
}
}
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