PHP - printf() prints part of my code

Question

When running this (simple) code:

<?php

$array = array("Blue", "Green", "Yellow", "Pink", "");

foreach ($array as $arrayElement) {
    printf("<div class = \"colorSubArea %s \" > <p> 1 </p> </div> " , $arrayElement);
}

https://jsfiddle.net/xbkky7jx/3/

the last part of my code is geting printed out. Also the loop doesn't continue after the first iteration (possibly the same cause).I am new in PHP and this is very confusing to me.

Perhaps because your webserver (like jsfiddle) isn't set up to execute PHP — Mark Baker, May 11 '16 at 17:49
Because JSfiddle doesn't do server side code. You need to test PHP code with a service that will execute the code on the server side. — Terry, May 11 '16 at 17:50
http://sandbox.onlinephpfunctions.com/code/058493d6ce9e2cef0b626c07a2e4494953b640c3 — AbraCadaver, May 11 '16 at 17:50
Apache absent of the `php` module would the same; serve the file without parser-logic. — , May 11 '16 at 17:51
You could use XAMPP to run your PHP code. It's a pretty neat tool and free. — Enigma, May 11 '16 at 17:51

Positivity · Accepted Answer · 2016-05-11T17:54:47.073

Your php code is ok, the reason you don't see the output is HTML, see the output in viewsource mode.

Your output is:

<div class = "colorSubArea Blue " > <p> 1 </p> </div> 
<div class = "colorSubArea Green " > <p> 1 </p> </div> 
<div class = "colorSubArea Yellow " > <p> 1 </p> </div> 
<div class = "colorSubArea Pink " > <p> 1 </p> </div> 
<div class = "colorSubArea  " > <p> 1 </p> </div>

see this: https://eval.in/568890

          $array = array("Blue", "Green", "Yellow", "Pink", "");

          foreach($array as $arrayElement){  
            printf("%s " , $arrayElement);
          }

PHP - printf() prints part of my code

1 Answers1