javascript - print numbers between 1- 20 using rules - Stack Overflow

admin管理员组

文章数量:1343373

I am going through codeacademys JavaScript tutorials as i am new to it. The tutorial asks for the following:

Print out the numbers from 1 - 20.
The rules:
- For numbers divisible by 3, print out "Fizz".
- For numbers divisible by 5, print out "Buzz".
- For numbers divisible by both 3 and 5, print out "FizzBuzz" in the console.
- Otherwise, just print out the number.

Here is my code:

for (i=1; i<=20; i++) {

if(i%3==0) {
    console.log("Fizz");
} 
if(i%5==0){
    console.log("Buzz");
}else if (i%5==0 && i%3==0) {
    console.log("fizzBuzz");
} else {
    console.log(i);
}

}

i am getting an error saying that i am printing out the wrong number of items, anyone know why that is?

I am going through codeacademys JavaScript tutorials as i am new to it. The tutorial asks for the following:

Print out the numbers from 1 - 20.
The rules:
- For numbers divisible by 3, print out "Fizz".
- For numbers divisible by 5, print out "Buzz".
- For numbers divisible by both 3 and 5, print out "FizzBuzz" in the console.
- Otherwise, just print out the number.

Here is my code:

for (i=1; i<=20; i++) {

if(i%3==0) {
    console.log("Fizz");
} 
if(i%5==0){
    console.log("Buzz");
}else if (i%5==0 && i%3==0) {
    console.log("fizzBuzz");
} else {
    console.log(i);
}

}

i am getting an error saying that i am printing out the wrong number of items, anyone know why that is?

• javascript

Share Improve this question Follow asked Oct 7, 2013 at 10:50 user2656943user2656943 3

• What is being output by your code? – Anthony Grist Commented Oct 7, 2013 at 10:54
• The output should be a list that prints out "Fizz" if the number is divisible by 3, "Buzz" if the number is divisible by 5 and "FizzBuzz" if the number is both divisible by 3 and 5 otherwise just print out the number – user2656943 Commented Oct 7, 2013 at 10:56
• It is checking for each line that you output. console.log adds a newline at the end of your output. So for numbers divisible by 15, it is printing Fizz and Buzz on separate lines. Also, your third code block can never be executed. (It will execute only if i%5==0 is false and i%5==0 is true; see the problem?) – Rohan Prabhu Commented Oct 7, 2013 at 10:56

Add a ment  | 

10 Answers 10

Sorted by: Reset to default 2

The check for both 3 and 5 must be first, otherwise the two other statements are true already and give separate logs. Now you print FizzBuzz in a single console.log statement.

   for (i=1; i<=20; i++) {

      if (i%5==0 && i%3==0) {
          console.log("FizzBuzz");
      } else if(i%3==0) {
          console.log("Fizz");
      } else if(i%5==0){
          console.log("Buzz");
      } else {
          console.log(i);
      }

    }

When the value is divisible by 3 and not by 5, on the first If statement "Fizz" is printed.

Then on the second if statement, the last else is hit so the number will also be printed. You will need to change the if(i%5==0) to else if.

However there will be a problem now when (i%5==0 && i%3==0) as the else if for that will never be hit. You can fix this by putting this as the first parison and changing the output to FizzBuzz.

Like this:

for ( i = 1; i <= 20; i++) {
    if (i % 5 === 0 && i % 3 === 0) {
        console.log("FizzBuzz");
    } else if (i % 3 === 0) {
        console.log("Fizz");
    } else if (i % 5 === 0) {
        console.log("Buzz");
    } else {
        console.log(i);
    }
};

Make sure you understand why this fixes your issue before you move on as you will most likely make the same mistake again.

Add a ment if you would like me to explain clearer if you are struggling to work out why you have gone wrong.

You should check if a number is divisible by both 3 and 5 in your first IF, since numbers that are divisible by both would otherwise result in the first and the second IF statement being executed.

Furthermore, the rules tell you to write out "FizzBuzz" in case a number is divisible by both, and at the moment you're only printing out "Buzz".

Did you try it out yourself? Because when you run it you get this output:

1
2
"Fizz"
3
4
"Buzz"
"Fizz"
6
7
8
"Fizz"
9
"Buzz"
11
"Fizz"
12
13
14
"Fizz"
"Buzz"
16
17
"Fizz"
18
19
"Buzz"

As you can see, you are printing out the number even when you printed Fizz and also you are actually supposed to print FizzBuzz in a single line, instead of two separate ones.

To fix the former issue you should take a look at your if/else structure. You have a separate if at the beginning just for Fizz. After that, you are handling the Buzz separately as well and if there’s no match for it, it will print out the number. So although you already printed Fizz you are still going to the last else. So you should bine those two separate if blocks into a single one.

The other issue is that console.log will always write a separate line. So in your case where you check for all the FizzBuzz conditions, you should print FizzBuzz. You will also want to check that first, otherwise the Buzz condition will hit first without giving you a chance to print FizzBuzz.

Try walking through the logic by hand. If i is 1, then only the last block works. If i is 3, the first block works. If i is 5, the second block works. And if i is 15, the second block works, and the third never gets a chance.

In general, have the most restrictive conditions run before the least restrictive. You should also notice and check carefully when two of your blocks are dissimilar. You have four blocks, and only one is an else block.

Here is I got the answer correct.

for(var i=1;i<=20;i++)
{

if(i%3 === 0 && i%5 === 0)
{
    console.log("FizzBuzz");
    }
else if(i%3 === 0)
{
    console.log("Fizz");
    }
    else if(i%5 === 0)
{
    console.log("Buzz");
    }
    else
    {
        console.log(i);
    }
}

The result will be:

1 2 Fizz 4 Buzz Fizz 7 8 Fizz Buzz 11 Fizz 13 14 FizzBuzz 16 17 Fizz 19 Buzz

Hi please check the ¨=¨

for (i = 1; i <= 20; i++) {
    if (i%3===0 && i%5===0) {
        console.log("FizzBuzz");
    } 
    else if (i%3===0) {
         console.log("Fizz");
    }
    else if (i%5===0) {
        console.log("Buzz");
    } 

    else {
        console.log(i);
    }
}

for (var i=1 ; i<=20 ; i++){
    if (i%3===0 && i%5!=0){
        console.log("Fizz");
    } else if (i%5===0 && i%3!=0){
        console.log("Buzz");
    } else if (i % 3 === 0 && i % 5 === 0 ){
        console.log("FizzBuzz");    
    } else {
        console.log (i);
    }
}

var i=1;
while (i<21)
{    
    if(i%3!==0 && i%5!==0)
    {
        console.log(i);
    }
    if(i%3===0 && i%5===0 )
    {
        console.log("FizzBuzz");
    }
    else if(i%3===0)
    {
        console.log("Fizz");
    }
    else if(i%5===0)
    {
        console.log("Buzz");
    }   

    i++;
}

I don't know Javascript, but I know some Python and I've e across the same problem. Maybe this can help. I'm sure someone can just translate this for Javascript.

Here's the original problem: Loop through the numbers between 1 and 20. If a number is divisible by 3, print Hip. If the number is divisible by 7, print “Hooray”.

Here's the code for python for a similar problem:

for numbers in range(1,21):
    if numbers % 3 !=0 and numbers % 7 != 0:
        print(numbers)
    if numbers % 3 == 0:
            print("Hip")
    if numbers % 7 == 0:
            print("Hooray")

Here's the output:

1 2 Hip 4 5 Hip Hooray 8 Hip 10 11 Hip 13 Hooray Hip 16 17 Hip 19 20

本文标签： javascriptprint numbers between 1 20 using rulesStack Overflow