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Sahan Balasuriya10,115 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 $$string; //display the corresponding excercise string ?>
Niki Molnar25,698 Points
These two lines are causing the problem:
$string = '$exercise' . "$day"; echo $$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 $$string // Error echo $string // Compiles
Paulo Dacaya32,900 Points
Maybe @Sahan Balasuriya was trying to achieve something like this.
$day = date('N'); $day = 6; $string = 'exercise' . "$day"; //$string holds the variable 'exercise6' echo $$string; //echo $exercise6 //result: Month long string of the day
Issue: Added an extra '$' in the program when assigning to $string.