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# 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
?>
```

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
```

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.