JavaScript Ternary with Logical OR

Question

var a = '1';
console.log(a == ('2'||'1')?'hi':'hello');

Doing so the condition get failed as a = 1 . It is comparing a's value 1 with 2 as this condition failed. so it always print hello. Is there any way to check for value('1') also which is after "||" so that it print hi?

Answer 1

var a = 1; a == '2'?'hi': a=='1'?'hello' :'';

i Found this is also working.... Thanks to everyone who answered using different ways..

Answer 2

Either list the different possibilities out separately:

a === '2' || a === '1' ? 'hi' : 'hello'

Or use an array and .includes:

['2', '1'].includes(a) ? 'hi' : 'hello'

The problem with ('2'||'1') is that the whole section there gets evaluated to a single expression before the comparison against a is made, and || will evaluate to the initial value if it's truthy. So ('2' || '1') resolves to '2'.