Question

我正在尝试解决以下问题（Udacity＆＃Javascript简介）：

路线：   写一个打印出以下歌曲的循环。从99开始，到1瓶结束。

示例：
     墙上挂着99瓶果汁！ 99瓶果汁！拿下一个，把它传递下去......墙上挂着98瓶果汁！
     墙上挂着98瓶果汁！ 98瓶果汁！拿下一个，通过它   在墙上...... 97瓶果汁！

...

墙上有2瓶果汁！ 2瓶果汁！拿一个，把它递过来......墙上有一瓶果汁！
     墙上挂了1瓶果汁！ 1瓶果汁！拿下一个，将它传递给...墙上的0瓶 s 果汁！

我的代码没有正确输出最后一行（它不包括＆＃34; s＆＃34;＆＃34;瓶＆＃34;）

我的代码如下所示：

 var num = 99;

 while (num >= 1) {

 num == 1 ? ((plural = "") && (nextPlural = "s")) :
 num == 2 ? ((plural = "s") && (nextPlural = "")) :
 ((plural = "s") && (nextPlural = "s"));

 console.log (num + " bottle" + plural + " of juice on the wall! " + num + "bottle" + plural + " of juice! " + "Take one down, pass it around... " + (num - 1) + " bottle" + nextPlural + " of juice on the wall!");
 num = num - 1
 }

为什么这段代码忽略了我的条件＆＃34; num == 2？＆＃34;在最后一行输出？

仅供参考，我能够使用以下代码解决这个问题，但它看起来并不干净，所以我想优化它：

var num = 99;
var plural = "s";
var nextNum = num - 1;
var nextPlural = "s";

while (num >= 1) {
if (num > 1 && nextNum > 1){
plural = "s";
nextPlural = "s";
}
else if (num > 1 && nextNum == 1){
plural = "s";
nextPlural = "";
}
else if (num == 1 && nextNum <= 1){
    plural = "";
    nextPlural = "s";
}
console.log(num + " bottle" + plural + " of juice on the wall! " + num + " bottle"+ plural + " of juice! " +
"Take one down, pass it around... " + nextNum + " bottle" + nextPlural + " of juice on the wall!");
num = num - 1;
nextNum = num - 1;
}

Answer 1

三元运算符的东西只是一团糟，我试图解决它，但它确实值得尝试修复，因为使用这样的三元运算符不是一个好主意。你已经在另一个三元内嵌了一个三元组，基本上有一个if, else if, else缩成3行。

你可以避免做额外的逻辑，并意识到当数字不是1时你只需要添加's'。

通过将此逻辑提取到返回字符串's'或空字符串''的函数中，您只需将此函数插入循环并为其提供n和{{ 1}}。

n-1

然后你可以简单地将你的循环改为

function plural(n) {
  return n == 1 ? '' : 's';  // note this is the appropriate usage of a ternary operator
}

Answer 2

好吧，问题是，如果我理解正确的话，为什么第一个代码示例的最后一行输出“0瓶”而不是“0瓶”。

因此，让我们将您的代码翻译为您想要实现的内容，并将您的代码翻译成英文并找出解释器的作用：

将num设为99。
当num大于或等于1时，请执行以下操作：

2.1如果num等于1，请将plural设为""，将nextPlural设为"s"。

2.2如果num等于2，则将plural设为"s"，将nextPlural设为""。

2.3否则将plural和nextPlural设置为"s"。
控制台输出很简单，所以我不会在这里提及。
将num设为num-1。

您可能会注意到用if语句替换三元运算符会产生正确的结果：

var num = 1;
var plural = '';
var nextPlural = '';

while (num >= 1) {

if(num==1) { plural = ''; nextPlural='s'; }

/*num == 1 ? ((plural = "") && (nextPlural = "s")) : (num == 2 ? ((plural = "s") && (nextPlural = "")) : ((plural = "s") && (nextPlural = "s")));*/

console.log (num + " bottle" + plural + " of juice on the wall! " + num + " bottle" + plural + " of juice! " + "Take one down, pass it around... " + (num - 1) + " bottle" + nextPlural + " of juice on the wall!");
num = num - 1

}

这是什么意思？您使用三元运算符时出现语法错误。它可能是什么？让我们声明一个新变量并尝试设置并将其与其他变量一起输出。

var num = 1;
var plural = '';
var nextPlural = '';
var test = '';

while (num >= 1) {

//if(num==1) { plural = ''; nextPlural='s'; }

num == 1 ? ((plural = "") && (nextPlural = "s") && (test = "test")) : (num == 2 ? ((plural = "s") && (nextPlural = "")) : ((plural = "s") && (nextPlural = "s")));

console.log (num + " bottle" + plural + " of juice on the wall! " + num + " bottle" + plural + " of juice! " + "Take one down, pass it around... " + (num - 1) + " bottle" + nextPlural + " of juice on the wall!");
console.log(test);
num = num - 1

}

您会注意到test仍然等于空字符串，就像nextPlural一样。这是因为using && is not the correct way of instantiating variables inside ternary constructions，所以这段代码将按预期工作：

var num = 99;
var plural = '';
var nextPlural = '';

while (num >= 1) {

//if(num==1) { plural = ''; nextPlural='s'; }

num == 1 ? (plural = "", nextPlural = "s") : (num == 2 ? (plural = "s", nextPlural = "") : (plural = "s", nextPlural = "s"));

console.log (num + " bottle" + plural + " of juice on the wall! " + num + " bottle" + plural + " of juice! " + "Take one down, pass it around... " + (num - 1) + " bottle" + nextPlural + " of juice on the wall!");
--num;

}

如果您有兴趣，我可能会为解决方案编程：

for(var i=99, w=' on the wall!'; i>0; i--) {
	console.log(returnEmpties(i)+w+' '+returnEmpties(i)+'! Take one down, pass it around... '+returnEmpties(i-1)+w);
}

function returnEmpties(n) { return n+' bottle'+(n==1?'':'s')+' of juice'; }

我的代码并没有认识到一个条件

2 个答案: