javascript - 未捕获的 SyntaxError:意外的标记 'var'
问题描述
我正在做这个练习,由于某种原因,我不断收到语法错误,当我将数字放入游戏时,它还显示“未捕获的 ReferenceError:检查未定义”我不确定我在哪里出错了。在这里感谢任何帮助,因为我是学习 javascript 的初学者。提前致谢。
这是我正在尝试创建游戏的代码笔:
https://codepen.io/Anozie/pen/qBOwBPq
以下是那些不想点击链接的人的代码和说明:
/* 说明:您将创建一个热游戏或冷游戏。游戏将从用户那里进行猜测(1 到 20 之间的数字),然后检查该猜测是否与 1 到 20 之间的随机生成的数字匹配。如果猜测与随机数匹配,则提醒用户他们中奖了并重置游戏。如果它很接近,请提醒他们它们很热,如果没有,请再试一次。用户应该猜测直到他们获胜,每次猜测后,如果它更接近,更温暖,如果它更远,更冷,则提醒他们。HTML 已为您提供。您只需要编写 javascript 部分。*/
//variables
var guess, answer, diffGuess;
var answer = 20;
var guess =
var diffGuess = answer - guess;
function submitguess();
document.queryselector(".guessInput").addEventListener('click', submit);
document.queryselector("#guessInput").textcontent = guess;
if (guess === answer){
console.log('You got it right');
}else if (diffGuess >= 1 && diffGuess <= 5){
console.log('You are cold');
}else if (diffGuess >= 6 && diffGuess <= 10){
console.log('You are colder');
}else if (diffGuess >= 11 && diffGuess <= 15){
console.log('You are lukewarm');
}else if (diffGuess >= 16 && diffGuess <= 19){
console.log('You are really warm');
};
这是HTML
<h2>Hot or Cold Game</h2>
<p>I've picked a random number between 1 and 20. Try to guess the number and I will tell you if you're getting hotter or colder!</p>
<form>
<input type="text" class="guess" >
<input type="submit" value="Submit guess" class="guessInput" onclick="return check()">
</form>
解决方案
我就是这样做的。
HTML 代码
<h2>Hot or Cold Game</h2>
<p>I've picked a random number between 1 and 20. Try to guess the number and I will tell you if you're getting hotter or colder!</p>
<form>
<input id="guess" type="number" class="guess" >
<input type="submit" value="Submit guess" class="guessInput" id="btn-submit" >
</form>
JS代码
var answer = 20;
function check(){
var guess = document.getElementById("guess").value;
guess = Number(guess);
var diffGuess = Math.abs(answer - guess);
if (guess === answer){
console.log('You got it right');
}else if (diffGuess >= 1 && diffGuess <= 5){
console.log('You are cold');
}else if (diffGuess >= 6 && diffGuess <= 10){
console.log('You are colder');
}else if (diffGuess >= 11 && diffGuess <= 15){
console.log('You are lukewarm');
}else if (diffGuess >= 16 && diffGuess <= 19){
console.log('You are really warm');
}
}
document.getElementById("btn-submit").addEventListener("click", function(event){
event.preventDefault();
check();
});
编辑 :- 您可以使用 Switch 语句代替一系列 if-else 条件,以获得更好的清晰度。
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