How to set variable in a function variable in php?

Question

I have a function variable like this... $aName = "My Name";

$sayHelloFunction = public function sayHello($aName){
       echo($aName);
    }

and I have something like this.....

callAFunctionFromFunction($sayHelloFunction);

Inside the "callAFunctionFromFunction", I do this:

            if(is_callable($sayHelloFunction)) { 
                  $sayHelloFunction(); 
        }

I found that the "My Name" can't display, what did I do wrong...

Community · Accepted Answer · 2017-05-23 11:55:50Z

2

I suggest you to look here and here as these exact threads deal with passing a function as a parameter. Also closures (anonymous functions) in PHP have no name (that's why they are called anonymous), so what you should do is something like that.

<?php $sayHelloFunction = function($aName){
   echo($aName);
};
if(is_callable($sayHelloFunction)) $sayHelloFunction("Testing 1,2,3");

edited May 23, 2017 at 11:55

CommunityBot

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answered May 1, 2011 at 7:34

Djumaka

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1 Comment

Felix Kling Over a year ago

Well, closures have been added in PHP 5.3. Only public is invalid here.

serby · Accepted Answer · 2011-05-01 07:32:27Z

2

The function sayHello expects a parameter $aName, but when you call it you don't pass in a value.

You would need to do this:

    if(is_callable($sayHelloFunction)) { 
              $sayHelloFunction("Hello John"); 
    }

Also you can't use the public access type with closures.

 $sayHelloFunction = function sayHello($aName) {
    echo($aName);
 }

edited May 1, 2011 at 7:32

answered May 1, 2011 at 7:26

serby

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