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Start your free trialPolo Moreno
11,782 PointsI need help in this challenge
this challenge say this.
Alter the 'elevatorCloseButton' function to follow the best practices in declaring variables within the scope of the function.
my answer is this
function elevatorCloseButton(pushed) {
var status = "open",
elevator;
name = "elevator"
if (pushed) {
var status = ("I'll close when I'm ready.");
}
}
elevatorCloseButton(false);
elevatorCloseButton(true);
but the system say this.
You should only need to do 'var status' once. I do not understand what is wrong and leave only "status", but still wrong can you help me please?
<!DOCTYPE html>
<html lang="en">
<head>
<title> JavaScript Foundations: Variables</title>
<style>
html {
background: #FAFAFA;
font-family: sans-serif;
}
</style>
</head>
<body>
<h1>JavaScript Foundations</h1>
<h2>Variables: Hoisting</h2>
<script>
function elevatorCloseButton(pushed) {
var status = "open",
elevator;
name = "elevator"
if (pushed) {
var status = ("I'll close when I'm ready.");
}
}
elevatorCloseButton(true);
elevatorCloseButton(true);
</script>
</body>
</html>
1 Answer
Justin Estrada
34,995 PointsDeclare the variable status at the beginning of the function then remove the declaration of the variable status within the If conditional block.
like this:
function elevatorCloseButton(pushed) { var status; if (pushed) { status = "I'll close when I'm ready."; } }