How to use function return value as parameter for another function in PHP?

Question

I want to have a function which takes 1 parameter as input, if no value is passed it should take default value as another function's output. Am using Class in PHP to achieve this but it's giving me error as "Constant expression contains invalid operations"

<?php

class Common
{
    public $usrname;

    function getLoggedInUser(){
        if (ISSET($_SESSION['email'])) {
            $this->usrname = ($_SESSION['email']);
            return $this->usrname;
        }
        return false;
    }

    function getLoggedInUserId($username = $this->usrname){
        echo $username;
    }

}

And calling the class file as

    <?php
    include "common.php"
    $c = new Common;
    $c->getLoggedInUserId();

Below is the error is shown for above call

Fatal error: Constant expression contains invalid operations

Please let me know how to pass function result as parameter for another function. Thanks

Nigel Ren · Accepted Answer · 2018-05-30 08:20:46Z

2

As a default value has to be a constant value, you would need to write something like this...

function getLoggedInUserId($username = null){
    if ( $username == null ) {
       $username = $this->usrname;
    }
    echo $username;
}

answered May 30, 2018 at 8:20

Nigel Ren

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Comments

Ramratan Gupta · Accepted Answer · 2018-05-30 08:22:28Z

1

getLoggedInUserId user must redefined as below

function getLoggedInUserId($username=""){
    if(empty($username)){
     echo $this->usrname;
    }else{
         echo $username;
    }

}

answered May 30, 2018 at 8:22

Ramratan Gupta

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Comments

samezedi · Accepted Answer · 2018-05-30 08:29:12Z

1

change your function to this:

public function getLoggedInUserId($username=""){
    $username = ($username == "") ? $this->usrname : $username;
}

answered May 30, 2018 at 8:29

samezedi

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1 Comment

Zeeshan Over a year ago

Thanks for the help. Got it :)

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