我的 JS 代码是否适合“墙上的 99 瓶啤酒”歌曲？ [关闭]

Question

JS新手在这里寻求一些建议。我一直在研究 Head First JavaScript Programming 这本书，它向我展示了如何将“墙上的 99 瓶啤酒”歌曲记录到控制台。这是原始代码：

var word = "bottles";
var count = 99;
while (count > 0) {
    console.log(count + " " + word + " of beer on the wall");
    console.log(count + " " + word + " of beer,");
    console.log("Take one down, pass it around,");
    count = count - 1;
    if (count > 0) {
        console.log(count + " " + word + " of beer on the wall.");
    } else {
        console.log("No more " + word + " of beer on the wall.");
    }
}

没有给出答案，它暗示代码没问题但不是 100% 正确，并询问您是否可以找到缺陷并​​修复它。在查看控制台后，我发现 while 循环没有考虑到一旦你在歌曲中找到一瓶啤酒，“bottles”这个词就需要变为“bottle”。我在循环中添加了一些 if 语句，以便在需要时将 var 单词更改为“bottle”。该代码现在似乎可以正常工作：

var word = "bottles";
var count = 99;
while (count > 0) {
  if (count == 1){
    var word = "bottle"
  }
    console.log(count + " " + word + " of beer on the wall");
    console.log(count + " " + word + " of beer,");
    console.log("Take one down, pass it around,");
    count = count - 1;
    if (count > 0) {
      if (count == 1){
        var word = "bottle"
      }
        console.log(count + " " + word + " of beer on the wall.");
    } else {
      if (count < 1){
        var word = "bottles"
      }
        console.log("No more " + word + " of beer on the wall.");
    }
}

我的问题是 - 有没有比我的尝试更好或更简洁的方法来做到这一点？我觉得我的尝试有点混乱。我知道您可以为此使用 For 循环，但该示例使用了 While 循环，所以如果可能的话，我想坚持下去。

非常感谢！

Answer 1

你可以通过用简洁的三元表达式替换那些笨重的 if else 语句来稍微整理一下你的逻辑：

while (count > 0) {
    var bottle = count == 1 ? "bottle" : "bottles";
    console.log(count + " " + word + " of beer on the wall");
    console.log(count + " " + word + " of beer,");
    console.log("Take one down, pass it around,");
    --count;
    var bottle = count == 1 ? "bottle" : "bottles";
    console.log(count + " " + word + " of beer on the wall.");
}

这在语法上假设短语0 bottles of beer on the wall 是正确的并且是我们想要使用的。这听起来对我来说是正确的，但我想从技术上讲，0 瓶或 1 瓶都应该是单数的。

Answer 2

由于您想紧贴示例，我只指出您可以跳过

if (count < 1){
    var word = "bottles"
  }

如果你改变了

 if (count == 1){
    var word = "bottle"
  }

进入

if (count == 1){
    word = "bottle"
  }

并在此处放置 var 声明：

while (count > 0) {
    var word = "bottles";

还有很多其他的东西你也可以缩短，但这可能会使它变得更加困难。

Answer 3

一种可能性是将整个瓶子单词的东西提取到一个单独的函数中。这样，您只需要更改原始代码中的任何内容

var count = 99;
while (count > 0) {
    console.log(count + " " + getBottleWord(count) + " of beer on the wall");
    console.log(count + " " + getBottleWord(count) + " of beer,");
    console.log("Take one down, pass it around,");
    count = count - 1;
    if (count > 0) {
        console.log(count + " " + getBottleWord(count) + " of beer on the wall.");
    } else {
        console.log("No more " + getBottleWord(count) + " of beer on the wall.");
    }
}

function getBottleWord(count) {
    return count === 1 ? "bottle" : "bottles";
}

Answer 4

// Instead of var you can start using 'const' for variables that will not be changed
// and 'let' for variables that will be changed.
let word = "bottles";
let count = 99;
while (count > 0) {
    if (count == 1){
        // No need to recall 'var' here, you are reassigning an existing var not creating a new one.
        // Can just do:
        // word = "bottle"
        word = "bottle"
    }
    /* As of ES6 You can use string templets here:
     *
     * console.log(`${count} ${word} of beer on the wall`);
     * console.log(`${count} ${word} of beer,`);
     * console.log("Take one down, pass it around,");
     */
    console.log(count + " " + word + " of beer on the wall");
    console.log(count + " " + word + " of beer,");
    console.log("Take one down, pass it around,");

    count = count - 1;
    /* For the word changing, you can lose some code by not actually changing the string,
     * That way you will not need to reassign it back to plural again when the count is 0
     * To take the 'bottles' string without the final plural 's' character use:
     * word.slice(0,-1);
     */
    if (count > 0) {
      if (count == 1){
        console.log(count + " " + word.slice(0, -1) + " of beer on the wall.");
      }
    } else {
        // now you can lost this if statment.
        console.log("No more " + word + " of beer on the wall.");
    }
}

Answer 5

您可以使用递归调用相同的函数，但瓶数更少，直到计数为 0

const pluralize = (i, word) => word + (i > 1 ? 's': '')
  
const bottles = count => {
  console.log(`${count} ${pluralize(count, 'bottle')} of beer on the wall.`)
  console.log(`${count} ${pluralize(count, 'bottle')} of beer.`)
  console.log(`Take one down, pass it around`)
  if (count > 1) {
    console.log(`${count - 1} ${pluralize(count - 1, 'bottle')} of beer on the wall.`)
    bottles(count - 1) // recursively call the bottles function with one less bottle
  }
  else {
    console.log(`No more bottles of beer on the wall.`)
  }
}

bottles(12)

Answer 6

好吧，您可以将 if-else 语句重新编写成更简洁的语句，例如：

var word = "bottles";
var count = 99;
while (count > 0) {
    console.log(count + " " + word + " of beer on the wall");
    console.log(count + " " + word + " of beer,");
    console.log("Take one down, pass it around,");
    count = count - 1;
    if (count > 1) {
        console.log(count + " " + count + " of beer on the wall.");
    } else if(count == 1) {
        console.log(count + " bottle of beer on the wall."); 
    } else {
        console.log("No more " + word + " of beer on the wall.");
    }
}

但在您的问题中值得指出的部分是：注意 多个 var word = "bottles" 句子，并且通常使用 var 关键字及其注意事项，尤其是关于它的范围。例如，考虑var 和let 与它们各自的scopes 之间的区别。

Answer 7

我是这样做的：

var word="bottles";
var count=99;
while (count>1) {
    console.log(count + " " + word + " of beer on the wall");
    console.log(count + " " + word + " of beer,");
    console.log("Take one down, pass it around,");
    count=count-1;
    
    if (count>1) {
        console.log(count + " " + word + " of beer on the wall.");
    }
    if (count==1) {
        console.log(count + " bottle" + " of beer on the wall.");
        console.log(count + " bottle" + " of beer on the wall,");
        console.log(count + " bottle" + " of beer");
        console.log("Take one down, pass it around,");
        count=count-1;
    }
    if (count==0) {
        console.log("No more " + word + " of beer on the wall.");
        
    }
}