CODEWITHSUNDEEP
javascript
INeedAlotOfHelpPleaseAndThanks
INeedAlotOfHelpPleaseAndThanks

Reputation: 21

if else statements being skipped, even if true

I have a basic table in javascript and a few lines of code where I compare the first values from each of the two tables. It skips the 'else if' statement and just goes straight to the 'else' at the end, when the 'else if' condition is true. I'm pretty new to all this so I won't be surprised if I messed up somewhere obvious. Any help is much appreciated.

var firstEquation = ['2', 'x', '+', '1', 'y', '=', '8'];
var secondEquation = ['3', 'x', '-', '1', 'y', '=', '7'];

if ( firstEquation[1] > secondEquation[1] ) {
    print("Outcome 1");
} else if ( firstEquation[1] < secondEquation[1] ) {
    print("Outcome 2");
} else {
    print("Else");
}

Upvotes: 1

Views: 1783

Answers (3)

McGlothlin
McGlothlin

Reputation: 2099

JavaScript starts counting array indexes at 0. You've essentially said:

if ( "x" > "x" ) {
    print("Outcome 1");
} else if ( "x" < "x" ) {
    print("Outcome 2");
} else {
    print("Else");
}

Since "x" is the second element in each array, and "x" = "x", you will always hit the else statement. Change your array indexes to firstEquation[0] and secondEquation[0] to compare the first elements of the arrays.

Upvotes: 2

Asaf Alster
Asaf Alster

Reputation: 26

You are comparing the second value, to compare first value use [0]. It goes straight to the last else because 'x' is not less nor greater than 'x'

Upvotes: 0

Shadow
Shadow

Reputation: 9427

Are you sure you want to use print? That's for sending data to paper.

It's possible you mean console.log - which would send that output to the console. If you came from a language such as python, then it would make sense that you didn't realise.

Upvotes: 1

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