I tried doing it little differently..

Question

So I tried to see if there is way to do this without using the if statement can someone check and tell me why this doesn't work. The console says undefined variable .

php
<?php
//store each excerice in a string variable
$exercise1 = 'display "Hello World!"';
$exercise2 = 'Convert Pounds to Kilograms';
$exercise3 = 'Convert Kilograms to Pounds';
$exercise4 = 'Convert Miles to Kilometers';
$exercise5 = 'Convert Kilometers to Miles';
$exercise6 = 'Month long string of the day';
$exercise7 = 'String of the day with levels';
//create a variable containing the day of the week
$day = date('N');
//use an if statement to test for the day of the week
$string = '$exercise' . "$day";
echo $$string;
//display the corresponding excercise string
?>

Answer 1 · 2017-08-28T21:00:51Z

August 28, 2017 9:00pm

These two lines are causing the problem:

$string = '$exercise' . "$day";
echo $$string;

You can't use variables inside single quotes, only in double quotes, so this is OK:

echo "Hello $name";

Whereas, the following would cause an error:

echo 'Hello $name';

To work with single quotes, you need to use concatenation:

echo 'Hello ' . $name;

Your line with echo $$string; uses two $$

echo $$string      // Error
echo $string       // Compiles

Answer 2 · 2018-02-15T07:47:39Z

February 15, 2018 7:47am

Maybe @Sahan Balasuriya was trying to achieve something like this.

$day = date('N');
$day = 6;

$string = 'exercise' . "$day"; 
//$string holds the variable 'exercise6'

echo $$string;
//echo $exercise6
//result: Month long string of the day

Issue: Added an extra '$' in the program when assigning to $string.

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Sahan Balasuriya

Sahan Balasuriya

I tried doing it little differently..

2 Answers

Niki Molnar

Niki Molnar

Paulo Dacaya

Paulo Dacaya