PHP using extract() to pass array values as variables and display it on a page

Question

I am attempting to create a view($path,$data) function in php the main functionality is include a specific file from the directory and pass in data/variable into that page, i managed to create $path and was able to include the define path, now my next step is pass in the $data values into my included page, and want to convert each array label as a variable.

My php class is below classes.php.

define("SITE_NAME", "process");

 class helpers {


        public function view($path, $data) 
        {

            $dir = $_SERVER['DOCUMENT_ROOT'].SITE_NAME."/";

            include($dir.$path.".php");

            return extract($data);



        }  

}

On my index.php

require_once('classes.php');

$helper = new Helpers();

$data['title'] = "My Test";
$data['test'] = "test1";

$helper->view('test',$data);

So now on my test.php i am tring to echo $title which i assuming it will return the value of My Test, to test if i am getting the values from index.php i was able to output the $data using print_r

Array ( [title] => My Test [test] => test1 )

Any advice how can i achieve? i was trying the extract() function but don't know if my syntax is correct.Thanks in advance!

i don't see any problem with your code. If there's error, i think its all good ;) — elegisandi
– elegisandi, Commented Aug 3, 2017 at 4:00
Yes $data.title would be good but i need to convert the title as a new variable which is $title. — KaoriYui
– KaoriYui, Commented Aug 3, 2017 at 4:03
Thanks for your response, is there anyway to convert $data['title'] into just $title so i can echo $title i was looking at the syntax of extract(array,extract_rules,prefix) i think i am missing the two parameters. — KaoriYui
– KaoriYui, Commented Aug 3, 2017 at 4:11
simple return the $data and in view simple apply the extract($data); — JYoThI
– JYoThI, Commented Aug 3, 2017 at 4:13
@JYoThI can you post a answer so i can reference it, it will help a lot thanks! — KaoriYui
– KaoriYui, Commented Aug 3, 2017 at 4:15

ArtisticPhoenix · Accepted Answer · 2017-08-03 05:03:55Z

4

Because return is limited to only one value.

By extracting it you are taking one value the array and splitting it into many.

Personally I avoid using things like extract and $$var in my code as it destroys my IDE and make readability near impossible. That said, in this case it does make sense to use it, because it's within a limited scope which limits the possibility of accidentally overriding another variable unintentionally.

http://php.net/manual/en/functions.returning-values.php

A function can not return multiple values, but similar results can be obtained by returning an array.

AND

http://php.net/manual/en/function.extract.php

extract — Import variables into the current symbol table from an array

symbol table ~ scope

Further

Return Values: Returns the number of variables successfully imported into the symbol table

public function view($path, $data)
{

    $dir = $_SERVER['DOCUMENT_ROOT'].SITE_NAME."/";

    include($dir.$path.".php");
    return extract($data); //returns the variables successfully imported from $data
}

.

As you can see once you call return, that ends the execution of the current function, and closes that scope. You will have to re-arrange these, so the variable assignment takes place first.

I'm assuming the included file in the method is something like this, it wasn't in the OP.

    <h1><?= $title; ?></h1>

You don't technically need to return anything because the HTML will naturally be caught by the output buffer and displayed when the script finishes executing. However, that is not very clean. The correct way is to take control of the output buffer like this:

    public function view($path, $data) 
    {
        $dir = $_SERVER['DOCUMENT_ROOT'].SITE_NAME."/";

        extract($data); //variables must exist before importing where they are used.
        ob_start();
        include($dir.$path.".php"); //capture this output
        $view = ob_get_clean();
       return $view;
    }  

   $out = $helper->view('test',$data); //out now contains the HTML output from the included file.

Then you can echo it out. IMO this is better because you could then insert that HTML output into another data array and pipe it into another template. Which is useful for things like a re-usable header or footer or nav bar etc..

Consider this

$head['page_title'] = "My Test";
$body['head'] = $helper->view('header',$head); //create head and assign to body

$body['name'] = 'John Smith';
echo $helper->view('body',$body); //create body (with head inserted) and echo

And header.php

<title><?= $page_title; ?></title>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js"></script>

And body.php

<!DOCTYPE html>
<html>
    <head>
        <?= $head; ?>
    </head> 
    <body>
       <p><?= $name; ?></p>
    </body> 
 </html>

The output would now be something like this

<!DOCTYPE html>
<html>
   <head>
      <title>My Test</title>
      <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js"></script>
  </head>   
  <body>
     <p>John Smith</p>
  </body>   
</html>

Now both pages are combined into one, that is output. You could take this to whatever level of re-usability you wanted, but it would save you a ton of typing and make maintaining the views way easier.

But, as I said you can just as simply allow it to output naturally

    public function view($path, $data) 
    {
        $dir = $_SERVER['DOCUMENT_ROOT'].SITE_NAME."/";

        extract($data); //variables must exist before importing where they are used.
        include($dir.$path.".php"); //capture this output
    } 
    //outputs when script is done.

edited Aug 3, 2017 at 5:03

answered Aug 3, 2017 at 3:59

ArtisticPhoenix

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19 Comments

Cornel Raiu Over a year ago

How is this supposed to help OP achieve what needed?

ArtisticPhoenix Over a year ago

@cornel.raiu - hold your horses

Cornel Raiu Over a year ago

Again. Explain why it is bad to use it that way and give a better solution :) you are only explaining why it.s bad

ArtisticPhoenix Over a year ago

it's not bad it makes no logical sense, structure wise its not possible to do that the scope is different, besides my wife just walked in and started yacking to me, so AS I SAID hold your horses. If you have a better solution your free to post your own answer, instead of jumping me..

Cornel Raiu Over a year ago

I am on my phone playing with my daughter. I can't post a solution now. But I would definitely post a solution that would actually help the OP correct the initial code

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