Passing a NULL value to a PHP function [duplicate]

Question

I have a function like:

function do_something($first, $second=null, $third=null) { 

    if ($isset($second)) {

        // Do something here

    }

}

Now, I want to pass a value for $third to the function, but I want $second to remain NULL:

do_something('abc','','def');

Now, $second is no longer NULL. How can I pass a value for $third while leaving $second NULL?

Well you could instead use the arguments such $first to signify mandatory and pass an$option = array() and proceed with the logic instead. But basically if u plan on providing that function with a lot of data. — Musk
– Musk, Commented Oct 8, 2015 at 16:47

Halayem Anis · Accepted Answer · 2015-10-08 16:42:58Z

4

There is no other way:

do_something('abc', null,'def');

answered Oct 8, 2015 at 16:42

Halayem Anis

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Comments

Community · Accepted Answer · 2017-05-23 11:59:20Z

2

I think it is quite easy to do this:

do_something('abc', null,'def');

Otherwise, I am linking this SO question which uses array arguments, in order to have optional arguments (credits to sanmai - I'd take no credit for this):

function do_something($arguments = array()) {
// set defaults
    $arguments = array_merge(array(
        "argument" => "default value", 
    ), $arguments); 

    var_dump($arguments);
}

With example usage:

do_something(); // with all defaults, or:
do_something(array("argument" => "other value"));

edited May 23, 2017 at 11:59

CommunityBot

11 silver badge

answered Oct 8, 2015 at 16:47

anon

1 Comment

Musk Over a year ago

My favored option, much cleaner, id use extract with that.

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