Javascript comparison

Question

well I want to know if my comparison has how to be improved, I want to make the code shorter (dry)

const teste = function(a){
   if(a != null && a != undefined && a != ""){
        console.log('e diferente')
    }else{
        console.log(`o valor que vc passou e ${a}`)
    }
}

teste()

About the part if(a != null && a != undefined && a != ""), how can I make it shorter?

Answer 1

Can only do:

if(a){
   ...
}else{
   ...
}

a would have to be different from null, Undefined and emptiness. Exactly the 3 conditions you put in your code, adding that also can’t be the number 0 (teste(0)). If you want to allow 0, would have to do:

if(a || a === 0){

Testing:

const teste = function(a){
   if(a){
        console.log('e diferente')
    }else{
        console.log(`o valor que vc passou e ${a}`)
    }
}

teste() // else
teste(' ') // if
teste(0) // else
teste('0') // if
teste('') // else

Answer 2

You can do it like this:

const teste = function(a) { a ? console.log(`O valor que vc passou é ${a}`) : console.log('Nenhum valor informado') };

teste('Teste');
teste('');
teste(null);
teste(undefined);

Or use Arrow Function

const teste = a => a ? console.log(`O valor que vc passou é ${a}`) : console.log('Nenhum valor informado');

teste('Teste');
teste('');
teste(null);
teste(undefined);

Answer 3

Other responses suggest a value check falsey, which excludes more than its condition.

Without changing the meaning of your original code, you can simply remove the first or second condition. They are identical because you are not using comparison of strict equality. So if the purpose of the code is to get smaller, you can simply use:

if (a != null && a != "")

If your intention with a != "" is to exclude only empty strings, use the strict equality comparator ===:

if (a != null && a !== "")

Otherwise it will also prohibit returning values '' when converts for strings - as empty arrays, for example.