How to use comparison operators

Question

For my JavaScript Course i am stuck on a question.

The question is:

On the editor to your right you find a variable named charmanderLevel, to which a value between 1 and 100 will be assigned.

Using else if statements print to the console which evolution of Charmander corresponds to that experience level. Consider an else statement if the experience level ever go above 100 that should print 'Charizard is as good as it gets '.

Here's a chart with the evolution which corresponds to each level:

Charmander - 1 to 15
Charmeleon - 16 to 35
Charizard - 36 to 100

The code i've build

var charmanderLevel = Math.ceil(Math.random() * 100);
if (charmanderLevel =>1) {
    console.log('Charmander');
} else if (charmanderLevel <=35) {
    console.log('Charmeleon');
} else if (charmanderLevel <=100) {
    console.log('Charizard');
} else {
    console.log('Charizard is as good as it gets');
}

With this code i get the warning:

Code is incorrect There should be one if statement checking if the variable charmanderLevel is greater than 1 and lesser than 15

I've tried to change my code to a value of <=15 but that gives the same warning. Also tried to make something to use >=1 and <=15 but that doesn't solve it either.

What goes wrong here and how to solve this?

It should be `charmanderLevel >= 1`, `=>` is for creating an arrow function — Hao Wu, Nov 19 '19 at 09:10
```charmanderLevel >= && charmanderLevel <= ``` in all if block should work. — Shail_bee, Nov 19 '19 at 09:13
And also as @Hao Wu noticed an equals sign(=) must always have a comparator sign(< or >) before it — pavan kumar, Nov 19 '19 at 09:16
A point to note for you would also be to make sure to not create where the if statement is always true and therefor the else statements will never be checked for example if you get the number 99 with your corrected function you'd get the output "Charmander" because it's greater than 1 and the other if's would therefor never be checked — tung, Nov 19 '19 at 09:45

score 0 · Answer 1 · answered Nov 19 '19 at 09:11

0

I guess thats correct

var charmanderLevel = Math.ceil(Math.random() * 100);
if (charmanderLevel >= 1 && charmanderLevel <= 15) {
    console.log('Charmander');
} else if (charmanderLevel >= 16 && charmanderLevel <= 35) {
    console.log('Charmeleon');
} else if (charmanderLevel <=100) {
    console.log('Charizard');
} else {
    console.log('Charizard is as good as it gets');
}

answered Nov 19 '19 at 09:11

Pawel Knuth

1

Solved! Made a little change to the Charizard numbers and now it works! – Richard Nov 19 '19 at 09:29
@Richard if you could leave upvote/solved would be nice :) – Pawel Knuth Nov 19 '19 at 10:25

score 0 · Answer 2 · answered Nov 19 '19 at 09:14

You could check only the upper bound, because the lower bound is one, as the description shows.

var charmanderLevel = Math.ceil(Math.random() * 100);

if (charmanderLevel <= 15) {
    console.log('Charmander');
} else if (charmanderLevel <= 35) {
    console.log('Charmeleon');
} else if (charmanderLevel <= 100) {
    console.log('Charizard');
} else {
    console.log('Charizard is as good as it gets');
}

score 0 · Answer 3 · answered Nov 19 '19 at 09:18

0

As the error says lesser than 15 not less than or equal to, I think it should be if(charLvl >= 1 && charLvl <15) Try this and let me know

answered Nov 19 '19 at 09:18

pavan kumar

823
6
15

score 0 · Answer 4 · answered Nov 19 '19 at 09:21

0

Try this, This will work.

var charmanderLevel = Math.ceil(Math.random() * 100);
if (charmanderLevel >= 1 && charmanderLevel <= 15) {
    console.log('Charmander');
} else if (charmanderLevel >= 16 && charmanderLevel <= 35) {
    console.log('Charmeleon');
} else if (charmanderLevel >= 36 && charmanderLevel <= 100) {
    console.log('Charizard');
} else {
    console.log('Charizard is as good as it gets');
}

answered Nov 19 '19 at 09:21

Idexigner

45
8

Yes this is the solution! – Richard Nov 19 '19 at 09:37

How to use comparison operators

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