Call a function with multiple default value in PHP

Question

How to call a function with multiple default value in PHP ?

I have a function like below

public function get( $action = 'find', $query = 'top' ) {
      //more code
}

What will happen if I call above function like below (with one value) ?

get('map') {
      //more code
}

Nothing, because your function contains no code. Suggestion: put `echo "action = $action, query = $query";` in your function, and try it. — KIKO Software, Jun 12 '22 at 09:57
What do you mean? Parameters are positional, unless you use named arguments in your function _call_. — Álvaro González, Jun 12 '22 at 10:06
What's unclear about this supposed behavior? What unexpected do you expect from this call? — Your Common Sense, Jun 12 '22 at 10:08
Read the friendly manual. https://www.php.net/manual/en/functions.arguments.php — Markus Zeller, Jun 12 '22 at 10:09
Thanks @ÁlvaroGonzález. How to use named arguments in function call ? — abu abu, Jun 12 '22 at 10:20
Look at [Example #9](https://www.php.net/manual/en/functions.arguments.php#example-163) and [named arguments](https://www.php.net/manual/en/functions.arguments.php#functions.named-arguments). — Álvaro González, Jun 12 '22 at 12:13

score 0 · Answer 1 · answered Jun 12 '22 at 10:06

0

You are using positional arguments. Passing only one argument in your function, becomes the value of first parameter i.e. $action The $query still has the value of 'top'.

answered Jun 12 '22 at 10:06

Sharjeel Ahmad

37
5

Thanks @Sharjeel Ahmad. How can I pass value for second parameter `$query` ? – abu abu Jun 12 '22 at 10:24

Call a function with multiple default value in PHP

1 Answers1