Javascript ternary operator result always the same

Question

The code below always responds with "hi". Even if i set the campaign1 variable to 0. Can anyone advise if I am writing this out incorrectly?

campaign1 = 2;

campaign_string = (typeof campaign1 > 1 ? "hello" : "hi" );

Kind regards,

Why are you using `typeof` to fetch the *value* of a variable? — Frédéric Hamidi, Jul 03 '15 at 13:53
Well `typeof campaign1` is always `"number"` regardless of which numerical value you give it. — JJJ, Jul 03 '15 at 13:53
In my code later on I am concatenating a string with the variable campaign_string. Is there a better way to do this? As i can't use an if statement in my string concat. — Amar, Jul 03 '15 at 13:55

score 1 · Answer 1 · answered Jul 03 '15 at 13:54

1

You don't need typeof operator. Try repeating without that; and get rid of the unnecessary paranthesis too.

This should do

campaign1 = 2;

campaign_string = campaign1 > 1 ? "hello" : "hi";

answered Jul 03 '15 at 13:54

score 1 · Answer 2 · answered Jul 03 '15 at 13:55

1

You are using typeof but not comparing it to a type. I think you mean to do this.

campaign1 = 2;

campaign_string = (campaign1 > 1 ? "hello" : "hi" );

typeof is used like this.

var variable = 'Test';

typeof variable === 'string' // This is true.

answered Jul 03 '15 at 13:55

Enijar

score 0 · Answer 3 · answered Jul 03 '15 at 13:54

0

You dont need this 'typeof', so change it to:

campaign1 = 2;

campaign_string = campaign1 > 1 ? "hello" : "hi" );

answered Jul 03 '15 at 13:54

Fabjan

3 Answers3