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PHP PHP Basics Unit Converter Manipulating Numbers

Vince Roszak
Vince Roszak
588 Points
Posted by Vince Roszak
Vince Roszak
588 Points

Cant figure why my code is not working. Please Help!!

The task is to: Create a New Float Variable named $floatOne with a value of 1.5. After adding 5 to $integerOne, and without further changes to the value of $integerOne or $floatOne, multiply $integerOne by $floatOne and display the results. The error is saying to use echo instead of using var_dump but I am using echo.

index.php 
<?php

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

$integerOne = $integerOne + 5;
var_dump(  $integerOne);

$integerTwo = $integerTwo - 1;
var_dump( $integerTwo);

$floatOne = 1.5;
echo $integerOne * $floatOne;

?>

1 Answer

Jessica Foster
Jessica Foster
6,950 Points
Jessica Foster
Jessica Foster
6,950 Points

Hi Vince,

Let's take this step by step.

The first task says: Create Two Integer Variables - The first will be named $integerOne with a value of 1 and the second will be named $integerTwo with a value of 2.

You did this correctly. Great job!

Ok, the next task says: Add 5 to $integerOne. Subtract 1 from $integerTwo.

This is where you got into a little trouble. You don't need to var_dump here. Simply perform the arithmetic operations and store them in the variables, like this:

<?php
$integerOne += 5; // I'm using combined operators, but you can also write it out like you did.
$integerTwo -= 1;
?>

The final task says: Create a New Float Variable named $floatOne with a value of 1.5. After adding 5 to $integerOne, and without further changes to the value of $integerOne or $floatOne, multiply $integerOne by $floatOne and display the results.

You also got this right. Once you get rid of the var_dump's earlier, you should be able to complete the challenge successfully.

If you have any more questions, feel free to ask!

Cheers, Jess