PHP function missing argument error

My validate function looks like that

function validate($data, $data2 = 0, $type)
{
...

Function call example

if ($result = validate($lname, 'name') !== true)
        response(0, $result, 'lname');

As you see, my validate function has 3 input vars. I’m not using second var – $data2 often, that’s why set it to 0 by default. But when I’m calling this function as given example (as far as I know it means $data=$lname, $data2=0, $type=’name’) getting error message

Missing argument 3 ($type) for validate() 

How can I fix that?

Answer

Missing argument 3 ($type) for validate() ^[1]

Always list optional arguments as the last arguments, never before non-optional arguments.

Since PHP doesn’t have named parameters¹ nor “overloading ala Java”, that’s the only way:

function validate($data, $type, $data2 = 0) {
}

---

• ¹ Error with severity E_WARNING until PHP 7.0 (including); Uncaught ArgumentCountError starting with PHP 7.1^rfc (and starting with PHP 8.0 as well for internal functions^rfc).

• ² before PHP 8.0, see Named Arguments