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PHP PHP Basics Daily Exercise Program Daily Exercise Program

Sahan Balasuriya
Sahan Balasuriya
10,115 Points

I tried doing it little differently..

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
?>

2 Answers

Niki Molnar
Niki Molnar
25,575 Points

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
Paulo Dacaya
Paulo Dacaya
32,900 Points

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.