Welcome to the Treehouse Community

Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community! While you're at it, check out some resources Treehouse students have shared here.

Looking to learn something new?

Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today.

Start your free trial

PHP PHP Basics Unit Converter Manipulating Numbers

Posted by phillip smith
phillip smith
1,681 Points

what am i missing here

what am i missing here

index.php 
<?php

//Place your code below this comment
$integerOne = 1;
$integerTwo = 2;
$floatOne = 1.5;

var_dump($integerOne + 5);
var_dump($integerTwo - 1);
?>

2 Answers

KRIS NIKOLAISEN
KRIS NIKOLAISEN
54,971 Points
KRIS NIKOLAISEN
KRIS NIKOLAISEN
54,971 Points

In general it is best to avoid var_dump in challenges. The error you receive is Bummer: You haven't changed $integerOne. The value of $integerOne after var_dump is still 1. Likewise $integerTwo is still 2. You need to update these so that after the calculation the assigned value of $integerOne is 6 and the assigned value of $integerTwo is 1.

phillip smith
phillip smith
1,681 Points

so what does the code look like to change those as you described?

KRIS NIKOLAISEN
KRIS NIKOLAISEN
54,971 Points
KRIS NIKOLAISEN
KRIS NIKOLAISEN
54,971 Points

You could do this:

<?php

//Place your code below this comment
$integerOne = 1;
$integerTwo = 2;
$floatOne = 1.5;

$integerOne += 5;
$integerTwo -= 1;

?>

which is short for this:

<?php

//Place your code below this comment
$integerOne = 1;
$integerTwo = 2;
$floatOne = 1.5;

$integerOne = $integerOne + 5;
$integerTwo = $integerTwo - 1;

?>