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，它表明代码没问题，但不是100％正确，并询问您是否可以找到该缺陷并修复它。在查看控制台后，我发现while循环没有考虑到“＃34; bottle＆＃34;需要改为＆＃34;瓶＆＃34;一旦你在歌曲中喝到一瓶啤酒。我已经在循环中添加了一些if语句，以使var字变为＆＃34; bottle＆＃34;需要的时候。代码似乎现在正常工作：
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呢？我有一种感觉，我的尝试有点凌乱。我知道你可以使用For循环，但是这个例子使用了While循环，所以我想在可能的情况下坚持使用。
非常感谢！
答案 0 :(得分：1)
你可以通过用简洁的三元表达式替换那些庞大的if else语句来整理你的逻辑：
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.");
}
这使得语法假设短语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，但我觉得从技术上来说，0或1瓶应该是单数。
答案 1 :(得分：0)
由于你想要贴近示例，我只会指出你可以跳过
if (count < 1){
var word = "bottles"
}
如果你改变了
if (count == 1){
var word = "bottle"
}
到
if (count == 1){
word = "bottle"
}
并将var声明放在这里：
while (count > 0) {
var word = "bottles";
还有很多其他事情你也可以缩短，但这可能会让它变得更加困难。
答案 2 :(得分：0)
一种可能性是将整个瓶子词的内容提取到一个单独的函数中。这样，你只需要改变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";
}
答案 3 :(得分：0)
好吧，你可以将if-else语句重写为更多succint，比如：
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.");
}
}
但值得指出的问题部分是：请注意多个var word = "bottles"句子，一般来说，使用var关键字及其警告，尤其是就其范围而言。例如，请考虑var和let及其各自scopes之间的差异。
答案 4 :(得分：0)
// 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} ${word} of beer,`);
* console.log("Take one down, pass it around,");
*/
console.log(count + " " + word + " of beer on the wall");
console.log(count + " " + word + " of beer,");
console.log("Take one down, pass it around,");
count = count - 1;
/* For the word changing, you can lose some code by not actually changing the string,
* That way you will not need to reassign it back to plural again when the count is 0
* To take the 'bottles' string without the final plural 's' character use:
* word.slice(0,-1);
*/
if (count > 0) {
if (count == 1){
console.log(count + " " + word.slice(0, -1) + " of beer on the wall.");
}
} else {
// now you can lost this if statment.
console.log("No more " + word + " of beer on the wall.");
}
}
答案 5 :(得分：0)
你可以使用递归来用更少的瓶调用相同的函数，直到你得到0计数
const pluralize = (i, word) => word + (i > 1 ? 's': '')
const bottles = count => {
console.log(`${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)
答案 6 :(得分：-1)
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.");
}
}