我的代码无法识别条件

Question

I am trying to solve this following problem (Udacity's Intro to Javascript):

Directions: Write a loop that prints out the following song. Starting at 99, and ending at 1 bottle.

Example:

99 bottles of juice on the wall! 99 bottles of juice! Take one down, pass it around... 98 bottles of juice on the wall!

98 bottles of juice on the wall! 98 bottles of juice! Take one down, pass it around... 97 bottles of juice on the wall!

...

2 bottles of juice on the wall! 2 bottles of juice! Take one down, pass it around... 1 bottle of juice on the wall!

1 bottle of juice on the wall! 1 bottle of juice! Take one down, pass it around... 0 bottle s of juice on the wall!

and my code doesn't appropriately output the last line (it doesn't include "s" after "bottle"):

My code looks like this:

 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
 }

why is this code ignoring my condition for "num == 2?" at the last line of output?

FYI, I was able to solve this using the following code, but it didn't look clean so I wanted to optimize this:

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

The ternary operator stuff is just a big mess, I've tried to figure it out, but it's really worth trying to fix since it's just not a good idea to use ternary operators like that. You've nested a ternary inside another ternary and basically have an if, else if, else condensed into 3 lines.

You can avoid doing extra logic and realize that you need only add an 's' when the number is not 1.

By extracting this logic into a function that returns the string 's' or the empty string '' , you can simply plug this function into your loop and provide it with n and n-1 .

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

You can then simply your loop to just

while (num >= 1) {
  console.log('...');  // left as an exercise for you to fill in the required function call and string concatenations.
  num--;  // same thing as num = num - 1;
}

Answer 2

Alright, so the question, if I understand it right, is why the last line for the first code sample outputs "0 bottle" instead of "0 bottles".

So let's translate what you attempted to achieve with your code into English and figure out what the interpreter does:

1. Set num to 99.

2. While num greater or equal to 1, do the following:

   2.1 If num equals 1, set plural to "" and nextPlural to "s" .

   2.2 Else if num equals 2, set plural to "s" and nextPlural to "" .

   2.3 Else set both plural and nextPlural to "s" .

3. The console output is trivial, so I won't mention it here.

4. Set num equal to num-1 .

You might notice that substituting the ternary operator with an if statement yields the correct result:

 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 } 

What does that mean? That there is a syntax error with your use of the ternary operator. What could it possibly be? Let's declare a new variable and try setting and outputting it along with the others.

 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 } 

You will notice that test remains equal to an empty string, just like nextPlural . That's because using && is not the correct way of instantiating variables inside ternary constructions , so this code will work as intended:

 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; } 

If you are interested, this is how I would probably program the solution:

 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'; }