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PHP

Posted by Elizabeth Bockmeyer
Elizabeth Bockmeyer
1,938 Points

I don't understand what I am doing incorrectly

<?php

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

//add 5 to intergerOne
$integerOne = $integerOne + 5;


//Subtract 1 from $integerTow
$integerTwo = $integerTwo - 1;


$floatOne = 1.5;

$integerOne = ($integerOne * $floatOne);

echo $integerOne;
?>
Steven Parker
Steven Parker
231,127 Points
Steven Parker
Steven Parker
231,127 Points

This looks like PHP code. Did you mean to put this question in the PHP category instead of JavaScript? If so, you can change the topic using "Edit Question".

Also, when posting code, please use the instructions for code formatting in the Markdown Cheatsheet pop-up below the "Add an Answer" area. :arrow_heading_down:

Iain Simmons
Iain Simmons
Treehouse Moderator 32,305 Points

Updated to change topic to PHP and add code in code blocks.

2 Answers

Steven Parker
Steven Parker
231,127 Points
Steven Parker
Steven Parker
231,127 Points

The trick is in the instructions.

You didn't provide a link, but I believe I have identified the challenge you are working on as Manipulating Numbers.

Task 3 of the challenge says: "... and without further changes to the value of $integerOne or $floatOne, multiply $integerOne by $floatOne and display the results.", but your code multiplies those numbers and assigns the result back to $integerOne. So instead of making that assignment, you should just echo the result of the multiplication itself.

Shaun Dixon
Shaun Dixon
10,944 Points
Shaun Dixon
Shaun Dixon
10,944 Points

I think the issue you are experiencing is being caused by the variables you are declaring, being on the same line as the comments.

Try this:

<?php

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

//add 5 to intergerOne 
$integerOne = $integerOne + 5;

//Subtract 1 from $integerTow 
$integerTwo = $integerTwo - 1;

$floatOne = 1.5;

$integerOne = ($integerOne * $floatOne);

echo $integerOne; ?>

Anything on the same line as the // will be classed as a comment and not parsed by PHP.

Steven Parker
Steven Parker
231,127 Points
Steven Parker
Steven Parker
231,127 Points

Note :point_right: Since Elizabeth did not use Markdown to format the code, that code might not actually be on the same line as the comments.

UPDATE: Now that Iain has formatted it, my suspicion has been confirmed.