PHP function working differently for return and echo. Why?

Question

By using the following class:

class SafeGuardInput{

    public $form;
    public function __construct($form)
    {
        $this->form=$form;
        $trimmed=trim($form);
        $specialchar=htmlspecialchars($trimmed);
        $finaloutput=stripslashes($specialchar);
        echo $finaloutput;
    }

    public function __destruct()
    {
        unset($finaloutput);
    }
}

and Calling the function, by the following code, it works fine.

        <?php 
        require('source/class.php');
        $target="<script></script><br/>";
        $forminput=new SafeGuardInput($target);
        ?>

But if in the SafeGuardInput class if I replace echo $finaloutput; with return $finaloutput; and then echo $forminput; on the index.php page. It DOES NOT WORK. Please provide a solution.

"It does not work" is not an adequate description of the problem, no matter how hard you hit capslock. — Sammitch
– Sammitch, Commented Jun 8, 2020 at 18:20
$finaloutput is only a local variable... unknown in the destructor — Honk der Hase
– Honk der Hase, Commented Jun 8, 2020 at 18:22

CaptHash · Accepted Answer · 2020-06-08 20:53:47Z

1

You can't return anything from a constructor. The new keyword always causes the newly created object to be assigned to the variable on the left side of the statement. So the variable you've used is already taken. Once you remember that, you quickly realise there is nowhere to put anything else that would be returned from the constructor!

A valid approach would be to write a function which will output the data when requested:

class SafeGuardInput{

    public $form;
    public function __construct($form)
    {
        $this->form=$form;
    }

    public function getFinalOutput()
    {
        $trimmed = trim($this->form);
        $specialchar = htmlspecialchars($trimmed);
        $finaloutput = stripslashes($specialchar);
        return $finaloutput;
    }
}

Then you can call it like in the normal way like this:

$obj = new SafeGuardInput($target);
echo $obj->getFinalOutput();

edited Jun 8, 2020 at 20:53

CaptHash

52 bronze badges

answered Jun 8, 2020 at 18:29

ADyson

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1 Comment

CaptHash Over a year ago

In the public function getFinalOutput, $form remains undefined variable.

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