How to pass specific variable in PHP function

function, parameters, php, variables

Solution

You cannot "not pass" a parameter that's not at the end of the parameters list :

- if you want to specify the 3rd parameter, you have to pass the 1st and 2nd ones

- if you want to specify the 2nd parameter, you have to pass the 1st one -- but the 3rd can be left out, if optionnal.

In your case, you have to pass a value for the 2nd parameter -- the default one, ideally ; which, yes, requires your to know that default value.

A possible alternative would be not have your function take 3 parameters, but only one, an array :

function my_function(array $params = array()) {
    // if set, use $params['first']
    // if set, use $params['third']
    // ...
}

And call that function like this :

my_function(array(
    'first' => 'plop',
    'third' => 'glop'
));

This would allow you to :

- accept any number of parameters

- all of which could be optionnal

But :

- your code would be less easy to understand, and the documentation would be less useful : no named parameters

- your IDE would not be able to give you hints on which parameters the function accepts

Problem

I have a PHP function that requires can take 3 parameteres... I want to pass it a value for the 1st and 3rd parameters but I want the 2nd one to default... How can I specify which ones I am passing, otherwise its interpreted as me passing values for the 1st and 2nd slots. Thanks.

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