How can I initialize an array's argument to its default value?

Question

Here is my code:

function myfunc( $arg1, $arg2 = 'sth', $arg3 = 'sth else' ){
    // do stuff
}

Now I want to set a new value for arg3 and keep the default defined value for arg2. How can I do that?

Something like this:

myfunc( true, <keep everything defined as default>, $arg3 = 'new value' );

The expected result is sth in this fiddle.

as per your `fiddle` you haven't defined `$arg2` before calling that function — Maitray Suthar, Jul 05 '17 at 06:05
your question is not clear please add some more explanation. — Yaseen Ahmad, Jul 05 '17 at 06:05
@YaseenAhmed I want to set a value for `arg3` without setting any value for `arg2`. As you see both `arg3` and `arg2` are optional. And I want to use the defined default value for `arg2`. — stack, Jul 05 '17 at 06:07
you can use array as argument and set the value inside function if not found. — rahul singh, Jul 05 '17 at 06:07
@SaadSuri nope .. As I said in the question, the expected result is `sth` in my fiddle. And in yours, the expected result is `array('sth','tth')`. — stack, Jul 05 '17 at 06:10
So you can make it array('sth'). unfortunately i can't eval this fiddle — Saad Suri, Jul 05 '17 at 06:12
I post the answer check it if you have any question let me know — Yaseen Ahmad, Jul 05 '17 at 06:16
@stack Please see this link https://stackoverflow.com/questions/10597114/php-default-function-parameter-values-how-to-pass-default-value-for-not-last — Saad Suri, Jul 05 '17 at 06:18

score 3 · Answer 1 · answered Jul 05 '17 at 06:05

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

function my_function(array $value = array()) {
    // if set, use $value['key1']
    // if set, use $value['key2']
    // ...
}

And call that function like this:

my_function(array(
    'key1' => 'google',
    'key2' => 'yahoo'
));

This would allow you to:

Hope it will helpful.

1 Answers1