Optional parameter in function php?

Question

i was having a look at a php script (namesilo module for whmcs) and found this function

function namesilo_transactionCall($callType, $call, $params)

and later in the script, it is called with the following code

namesilo_transactionCall("getNameServers", $apiServerUrl . "/api/getDomainInfo?version=1&type=xml&key=$apiKey&domain=$sld.$tld");

When it is called there is only two parameters and when the function was declared, the $params was not made optional. How is it possible. I am relatively new to php

I think it's not possible! something is wrong

Alireza Fallah
– Alireza Fallah

2014-01-26 19:25:42 +00:00
Commented Jan 26, 2014 at 19:25 — Alireza Fallah
– Alireza Fallah, Commented Jan 26, 2014 at 19:25

voodoo417 · Accepted Answer · 2014-01-26 19:57:27Z

1

I saw @Alireza Fallah comment, and decided to answer . Why it`s impossible?

 <?php 

  function namesilo_transactionCall($callType, $call, $params){
    var_dump($callType);
  }

  namesilo_transactionCall(1,1);
  // OUTPUT
  1

 ?>

If you doesnt pass defined not-optional agrument - interpreter generate Warning like:

Warning: Missing argument 3 for namesilo_transactionCall()

Set error_reporting(0) - and you never see this. You can ,in general, not specify all parameters like as here:

 function A(){  
   print_r(func_get_args());
 }

 // and call 

 A(1,2,3,4,5,6,7,8....N);

And this code will work. See more about functions in PHP.

edited Jan 26, 2014 at 19:57

answered Jan 26, 2014 at 19:48

voodoo417

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