未捕获的 SyntaxError：意外的标记 'var'

Question

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。 在此感谢任何帮助，因为我是学习 javascript 之旅的初学者。 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。 游戏将从用户那里得到一个猜测（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>

Answer 1

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：- 而不是一系列 if-else 条件，您可以使用 Switch 语句以获得更好的清晰度。