Pass values to specific arguments in PHP

Question

If I have a function such as

function foo($a, $b = 2, $c = 3)
{
  return [$a, $b, $c];
}

Then I can omit passing $b and $c since they have default values. But, say, I want to pass something to $c, but not to $b. How can I do it? Obviously I could swap their places in the declaration (function foo($a, $c = 3, $b = 2), but then I wouldn't be able to only pass $b.

My point is that I want to make a function that has arguments with default values, but which ones I want to pass varies. I tried this: print_r(foo(1, $c = 5)), but it results in [1, 5, 3] not in [1, 2, 5].

Also http://stackoverflow.com/questions/1342908/named-php-optional-arguments — Jonathon Reinhart, Feb 26 '17 at 15:12
You can't skip the second parameter. If you need to pass the third parameter you *must* pass the correct value for the second paramter. — John Conde, Feb 26 '17 at 15:12
Not possible. What I do is : `foo($param1, 2, $param3)` i.e. preserve the default. — apokryfos, Feb 26 '17 at 15:12

ScaisEdge · Accepted Answer · 2017-02-26T15:15:29.230

2

PHP does not support named parameters. You can use only ordered parameters with proper valued in right position

eg

foo( $a, 4 , 3);

or eventually unassigned value became null in related vars

   foo( $a,  null , 3);

result

 esults in [1, null, 3]

edited Feb 26 '17 at 15:15

answered Feb 26 '17 at 15:10

ScaisEdge

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This is incorrect. NULL will become the second value in that array. https://3v4l.org/TvvQP – John Conde Feb 26 '17 at 15:11
Okay, thanks. This looks hideous, but that's not the only un-aesthetical thing in php, I guess. – Fiodor Feb 26 '17 at 15:13
@JohnConde cirrect answer updated – ScaisEdge Feb 26 '17 at 15:13
Still incorrect: https://3v4l.org/vHQoe – John Conde Feb 26 '17 at 15:14

Pass values to specific arguments in PHP

1 Answers1