Representación condicional en React usando Ternaries y Logical AND

Hay varias formas en que su componente React puede decidir qué renderizar. Puede utilizar la ifdeclaración tradicional o la switchdeclaración. En este artículo, exploraremos algunas alternativas. Pero tenga en cuenta que algunos vienen con sus propias trampas, si no tiene cuidado.

Ternario vs if / else

Digamos que tenemos un componente al que se le pasa un nameprop. Si la cadena no está vacía, mostramos un saludo. De lo contrario, le decimos al usuario que debe iniciar sesión.

Aquí hay un componente de función sin estado (SFC) que hace precisamente eso.

const MyComponent = ({ name }) => { if (name) { return ( Hello {name} ); } return ( Please sign in );};

Muy claro. Pero lo podemos hacer mejor. Aquí está el mismo componente escrito usando un operador ternario condicional .

const MyComponent = ({ name }) => ( {name ? `Hello ${name}` : 'Please sign in'} );

Observe lo conciso que se compara este código con el ejemplo anterior.

Algunas cosas a tener en cuenta. Debido a que estamos usando la forma de declaración única de la función de flecha, la returndeclaración está implícita. Además, el uso de un ternario nos permitió SECAR el duplicado

o"> markup. ?

Ternary vs Logical AND

As you can see, ternaries are wonderful for if/else conditions. But what about simple if conditions?

Let’s look at another example. If isPro (a boolean) is true, we are to display a trophy emoji. We are also to render the number of stars (if not zero). We could go about it like this.

const MyComponent = ({ name, isPro, stars}) => ( Hello {name} {isPro ? '?' : null} {stars ? ( Stars:{'⭐️'.repeat(stars)} ) : null} ); 

But notice the “else” conditions return null. This is becasue a ternary expects an else condition.

For simple if conditions, we could use something a little more fitting: the logical AND operator. Here’s the same code written using a logical AND.

const MyComponent = ({ name, isPro, stars}) => ( Hello {name} {isPro && '?'} {stars && ( Stars:{'⭐️'.repeat(stars)} )} ); 

Not too different, but notice how we eliminated the : null (i.e. else condition) at the end of each ternary. Everything should render just like it did before.

Hey! What gives with John? There is a 0 when nothing should be rendered. That’s the gotcha that I was referring to above. Here’s why.

According to MDN, a Logical AND (i.e. &&):

expr1 && expr2Returns expr1 if it can be converted to false; otherwise, returns expr2. Thus, when used with Boolean values, && returns true if both operands are true; otherwise, returns false.

OK, before you start pulling your hair out, let me break it down for you.

In our case, expr1 is the variable stars, which has a value of 0. Because zero is falsey, 0 is returned and rendered. See, that wasn’t too bad.

I would write this simply.

If expr1 is falsey, returns expr1, else returns expr2.

So, when using a logical AND with non-boolean values, we must make the falsey value return something that React won’t render. Say, like a value of false.

There are a few ways that we can accomplish this. Let’s try this instead.

{!!stars && ( {'⭐️'.repeat(stars)} )}

Notice the double bang operator (i.e. !!) in front of stars. (Well, actually there is no “double bang operator”. We’re just using the bang operator twice.)

The first bang operator will coerce the value of stars into a boolean and then perform a NOT operation. If stars is 0, then !stars will produce true.

Then we perform a second NOT operation, so if stars is 0, !!stars would produce false. Exactly what we want.

If you’re not a fan of !!, you can also force a boolean like this (which I find a little wordy).

{Boolean(stars) && (

Or simply give a comparator that results in a boolean value (which some might say is even more semantic).

{stars > 0 && (

A word on strings

Empty string values suffer the same issue as numbers. But because a rendered empty string is invisible, it’s not a problem that you will likely have to deal with, or will even notice. However, if you are a perfectionist and don’t want an empty string on your DOM, you should take similar precautions as we did for numbers above.

Another solution

A possible solution, and one that scales to other variables in the future, would be to create a separate shouldRenderStars variable. Then you are dealing with boolean values in your logical AND.

const shouldRenderStars = stars > 0;
return ( {shouldRenderStars && ( {'⭐️'.repeat(stars)} )} );

Then, if in the future, the business rule is that you also need to be logged in, own a dog, and drink light beer, you could change how shouldRenderStars is computed, and what is returned would remain unchanged. You could also place this logic elsewhere where it’s testable and keep the rendering explicit.

const shouldRenderStars = stars > 0 && loggedIn && pet === 'dog' && beerPref === 'light`;
return ( {shouldRenderStars && ( {'⭐️'.repeat(stars)} )} );


I’m of the opinion that you should make best use of the language. And for JavaScript, this means using conditional ternary operators for if/else conditions and logical AND operators for simple if conditions.

While we could just retreat back to our safe comfy place where we use the ternary operator everywhere, you now possess the knowledge and power to go forth AND prosper.

I also write for the American Express Engineering Blog. Check out my other works and the works of my talented co-workers at You can also follow me on Twitter.