php - using optional arguments in php

Question

Suppose I have a function defined as following :

public function test($title, $body, $role = 0, $place = 0)
{
    //do something
}

so, when i call the function like this :

test('myname', 'sometext', third_arg);

I want 'third_arg' to be the argument for $place and no argument for $role.
I also don't know which of the $role or $place will have a value passed when the function is called everytime.
How do I pass the argument for $place ? Thanks in advance!

Jayesh Chitroda · Accepted Answer · 2016-07-05 10:14:48Z

2

Pass 0 OR null in 3rd argument:

test('myname', 'sometext', 0 ,third_arg);

OR

test('myname', 'sometext', '' ,third_arg);

edited Jul 5, 2016 at 10:14

answered Jul 5, 2016 at 10:00

Jayesh Chitroda

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2 Comments

Thamilhan Over a year ago

better to send 0 instead of empty string. Since the default value is set to 0

node_modules Over a year ago

.. Or you can also use the boolean false

Vimlesh · Accepted Answer · 2016-07-05 10:12:52Z

0

As the default value for third argument is 0 just pass the value 0 to the third. e.g: test('myname', 'sometext', 0,third_arg) ;

answered Jul 5, 2016 at 10:12

Vimlesh

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