Question

How to exclude a variable from being required in a function?

IE:

function foo($name,$address,$pizza_preference,$date)
{
if(!$pizza_preference)
{
return array($name,$address,$date);    
}
else
{
return array($name,$address,$pizza_preference,$date);
}
}

When calling this function how would I set it up so $pizza_preference is not required, but optional? So that if you only entered 3 arguments in the function it omits $pizza_preference, or would I have to make it so when you enter 0 it just doesn’t return it?

Answer 1

Just define a default value for it. Then you can use that function without passing a value:

function foo($name,$address,$date,$pizza_preference=null)
{
    if(!$pizza_preference)
    {
        return array($name,$address,$date);    
    }
    else
    {
        return array($name,$address,$pizza_preference,$date);
    }
}

Usually you put variables that have default values at the end of the parameters list so you don’t have to include blank parameters when calling the function.

See Default argument values on the PHP website for more.

UPDATE

If you’re going to have multiple parameters with default values and want to be able to skip them individually you can pass an array as the only parameter and read the values from there:

function foo(array $parameters)
{
    if(!$parameters['pizza_preference'])
    {
        return array($parameters['name'],$parameters['address'],$parameters['date']);    
    }
    else
    {
        return array($parameters['name'],$parameters['address'],$parameters['date'],$parameters['pizza_preference']);  
    }
}