PHP Optional Parameters - specify parameter value by name?

Question

I know it is possible to use optional arguments as follows:

function doSomething($do, $something = "something") {

}

doSomething("do");
doSomething("do", "nothing");

But suppose you have the following situation:

function doSomething($do, $something = "something", $or = "or", $nothing = "nothing") {

}

doSomething("do", $or=>"and", $nothing=>"something");

So in the above line it would default $something to "something", even though I am setting values for everything else. I know this is possible in .net - I use it all the time. But I need to do this in PHP if possible.

Can anyone tell me if this is possible? I am altering the Omnistar Affiliate program which I have integrated into Interspire Shopping Cart - so I want to keep a function working as normal for any places where I dont change the call to the function, but in one place (which I am extending) I want to specify additional parameters. I dont want to create another function unless I absolutely have to.

+1 to @sanmai; also note that Perl, Python, and Ruby all support named parameters in the way you describe. If this feature is important enough, you should consider using one of those languages. — Bill Karwin
– Bill Karwin, Commented Oct 27, 2010 at 2:42

sanmai · Accepted Answer · 2013-05-09 01:07:52Z

12

No, in PHP that is not possible as of writing. Use array arguments:

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

    var_dump($arguments);
}

Example usage:

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

When changing an existing method:

//function doSomething($bar, $baz) {
function   doSomething($bar, $baz, $arguments = array()) {
    // $bar and $baz remain in place, old code works
}

edited May 9, 2013 at 1:07

answered Oct 27, 2010 at 1:59

sanmai

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3 Comments

ClarkeyBoy Over a year ago

This seems like the kind of thing I should use. Will tick it as the correct answer when I am able to. Thanks.

ClarkeyBoy Over a year ago

Just thought of one thing - currently there are 4 arguments, two of which are optional. I want to add another 4. By implementing this method I would be reducing it to one argument (an array). This would break any calls to the function outside of the file I want to alter this in. Its been ages since I last did any PHP, but I am assuming that I could just create another function called the same thing but with a different number of arguments. Not ideal - but that should work.

sanmai Over a year ago

@ClarkeyBoy You can have as many arguments as you wish. See additional example.

Community · Accepted Answer · 2023-11-14 22:25:35Z

3

Have a look at func_get_args: https://www.php.net/manual/en/function.func-get-args.php

edited Nov 14, 2023 at 22:25

CommunityBot

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answered Oct 27, 2010 at 1:57

zerodin

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1 Comment

ClarkeyBoy Over a year ago

Thanks for the input - but I think that sanmai's answer suits my situation a bit better. With this method I would not be able to retrieve an argument by name - so I would hit the same brick wall (I think).

Jason McCreary · Accepted Answer · 2010-10-27 01:59:40Z

2

Named arguments are not currently available in PHP (5.3).

To get around this, you commonly see a function receiving an argument array() and then using extract() to use the supplied arguments in local variables or array_merge() to default them.

Your original example would look something like:

$args = array('do' => 'do', 'or' => 'not', 'nothing' => 'something');
doSomething($args);

answered Oct 27, 2010 at 1:59

Jason McCreary

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Comments

mario · Accepted Answer · 2010-10-27 02:01:23Z

0

PHP has no named parameters. You'll have to decide on one workaround.

Most commonly an array parameter is used. But another clever method is using URL parameters, if you only need literal values:

 function with_options($any) {
      parse_str($any);    // or extract() for array params
 }

 with_options("param=123&and=and&or=or");

Combine this approach with default parameters as it suits your particular use case.

answered Oct 27, 2010 at 2:01

mario

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2 Comments

sanmai Over a year ago

What an overkill! Consider passing an object this way.

mario Over a year ago

@sanmai: Always depends on the usage. Microoptimizations are for fools. And imaginary overhead is preferable to hindering yourself with a clumsy API.

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