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Dwight Schofield
2,686 Pointswhat's wrong with my answer?
i get a message saying that there is a parse error. i remember having to use "parseInt " to convert a number in the string form to a number that's not in string form. but i don't know how to use "parse" in other situations.
do {
var secret = prompt("What is the secret password?");
} while ( secret !== "sesame" ) {
document.write("You know the secret password. Welcome.");
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<title>JavaScript Loops</title>
</head>
<body>
<script src="script.js"></script>
</body>
</html>
2 Answers
andren
28,558 Points"Parse error" in this instance refers to the fact that JavaScript had issues parsing your code, meaning that there is something wrong with how your code is laid out or the syntax you use that prevents it from properly understanding what you are trying to do.
The issue that is causing the parsing error is that you have an opening curly brace { after the while statement which does not belong there. If you remove it then the parse error goes away.
There is still another issue with your code though so it still won't pass after fixing that issue. The problem is that with the way your code is written the secret variable is redeclared each time the loop runs, which is quite inefficient.
It is better to declare secret outside the loop, and then just change the value stored within it in the loop (remember that you don't use the var keyword when changing a variable, only when declaring it). Once you have done that you will be able to pass the challenge.
Marshall Wells
8,368 PointsThis helped me lol - awesome
Dwight Schofield
2,686 Pointsthanks Andren. very helpful...
Grace Ji
5,402 PointsThis works :)
var secret;
do {
secret = prompt("What is the secret password?");
} while ( secret !== "sesame")
document.write("You know the secret password. Welcome.");
Marshall Wells
8,368 PointsMarshall Wells
8,368 PointsA parse error means your code doesn't follow syntax. In the code above you have not closed your While loop....... you need a '}' after '...Welcome.");'