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David Enriquez
7,983 PointsWhy does task 1 keep not passing when I am leaving the code from the previous task in place?
JavaScript getYear task 3 out of 3
function getYear (){
var year = new Date().getFullYear();
var yearToday = return year;
}
<!DOCTYPE HTML>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<title>JavaScript Basics</title>
</head>
<body>
<script src="script.js"></script>
</body>
</html>
2 Answers

Ryan Ruscett
23,307 PointsThis is because when you finish task 1. It checks task 1. When you go to task 2. It runs the checks for task 1 and 2. When you do task thee. It runs the task for 1, 2 and 3. But in the event where there is a compiler issue. It can't compile the code and thus not run any of the test. SO it just says test 1 failed. Cause it did, sine it couldn't compile the page to run test 1, 2 in order to get to three.
BUT you are soooo close.
You return year but you do not return the variable to a variable inside the same function.
Like this.
unction getYear (){
var year = new Date().getFullYear();
return year
}
yearToday = getYear();

rydavim
18,786 PointsThe return statement should be on it's own line, as the last thing in the function.
function getYear() {
var year = new Date().getFullYear();
return year;
}
If you've gotten past that bit and on to task three, your call to the function should be outside of it completely. So whatever line you write for step three should go after your code from the other steps.
function getYear() {
var year = new Date().getFullYear();
return year;
}
// Step three goes down here, please!
// Remember that you can set a variable equal to a function call and it will hold the return value!

David Enriquez
7,983 PointsAwesome, thank you for the quick reply!
David Enriquez
7,983 PointsDavid Enriquez
7,983 PointsThank you Ryan!
I forgot you can simply call a function in a variable.