我认为这是一个范围问题，因为我尝试在userPrmpt内设置我的函数firstPartCook并且它有效。我把它放在外面，但事实并非如此。所以它正在阅读，但不保留返回的内容。我想通过将它放在测试变量中它可以工作，但事实并非如此。
这是我的代码
HTML
<!DOCTYPE html>
<html>
<head>
<title> Race Makin' </title>
<!-- Link to the JS portion for this script -->
<script src="riceMakin.js"></script>
</head>
<body>
<!-- not going for anything fancy. Just focusing on mechanics -->
<h1>Lets cook some damn rice</h1>
<h2> To start cooking Rice, please hit start</h2>
<button onclick="firstPartCook()">Start</button>
<h3>Below explains the status on the Rice Making Machine:</h3>
<!-- with JS i have what is inbetween the spans, switching from on and off -->
<p id="print">Status: Turned <span id="print">Off</span> </p>
</body>
</html>
JS
//Global Vars here
//Promp for the User to continue throught the script. To use what is returned, set to a var
function userPrmpt(userResponse) {
prompt(userResponse);
}
// This function is for adding type to the DOM.
function insertCopy (copy) {
var paragraphCreate = document.createElement("p");
var copyNode = document.createTextNode(copy);
paragraphCreate.appendChild(copyNode);
var idSelecter = document.getElementById("print");
//print is the id tag of the span
idSelecter.appendChild(paragraphCreate);
}
//This is where we start working on the mechanics of the script
function firstPartCook() {
//var userAnswer = prompt("Welcome to the Rice Maker, want to start making rice?");
var test = userPrmpt("Hello");
if (test == "yes") {
console.log(test);
insertCopy("It worked");
} else {
console.log(test);
insertCopy("Nope");
}
}
答案 0 :(得分：0)
你必须从提示符返回值，否则函数将只返回undefined，这是任何没有返回其他值的函数的默认返回值
function userPrmpt(userResponse){
return prompt(userResponse);
}