PHP function missing argument error

Question

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 1

Missing argument 3 ($type) for validate()

Always list optional arguments as the last arguments. Since PHP doesn't have named parameters nor "overloading ala Java", that's the only way:

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

Answer 2

You should at least set the $type in this line:

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

at NULL or '' as you can see here:

function validate($data, $data2 = 0, $type = null)

PHP let you to set a value for the parameters, but you can't define a parameter WITHOUT a preset value AFTER parameter(s) which HAVE a preset value. So if you need to always specify the third param, you have to switch the second and the third like this:

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