React inline conditional component attribute

Question

I've been searching everywhere and can't find an answer to my question. So I want a conditional attribute which is only displayed on certain conditions, example:

<Button {this.state.view === 'default' && 'active'}></Button>

As you can see I only want to indicate the button active if the this.state.view is equal to default. However, I get Unexpected token, error...

But when I try to put an attribute before it, for example:

<Button isActive={this.state.view === 'default' && 'active'}></Button>

It passes the syntax error and displays fine but this is not what I wanted to achieve.

How can I fix this? And what could be the reason behind it not passing?

UPDATE

So I just found out that in react-bootstrap the property active is a shorthand of active=true so I solved it using

<Button active={this.state.view === 'default'}></Button>

So In case, someone encounters this problem I'm leaving it here. However, I still want to know why the conditional attribute is failing without enclosing it inside a prop-like syntax, example:

This:

active={this.state.view === 'default'}

Versus

{this.state.view === 'default' && 'active'}

Answer 1

First of all, JSX is just a syntactic sugar for React.createElement. So, it may look like, but, in reality, you don't specify html attributes: in fact, you are always passing props.

For instance, the JSX code <input type="button" value="My button" /> is transpiled into React.createElement('input',{type:'button',value:'My Button'}). And, when necessary, React engine renders this React Element to the DOM as a HTML element.

That said, we have that JSX transpiles a prop without a value as true (check docs). For instance: <Button active></Button> is transpiled to React.createElement(Button, { active: true });.

But, we know that HTML5 specification does not accept attribute=true (as pointed here). For instance: <button disabled=true></button> is invalid. The correct is <button disabled></button>.

So, to render the HTML element to the DOM, React considers only props that are valid attributes (if not, the attribute is not rendered). Check all supported html attributes. And, then, finally, if it's a boolean attribute, it removes the true/false value, rendering it properly.

For instance: <button active={true}></button> is transpiled to React.createElement("button", { active: true }); and then React renders to the DOM <button></button>, because there is no active attribute in HTML specification for the <button/> tag (is not in the supported attributes list).

But <button disabled={true}></button> is transpiled to React.createElement("button", { disabled: true }); and React renders to the DOM <button disabled></button>.

I just said that to clarify your case.

You're trying to pass an active prop to the Button component (first letter uppercase means that's a React component: there is a React Component called Button handled somewhere in your code).

That means:

<Button active></Button> is transpiled to React.createElement(Button, { active: true });

and

<Button active={true}></Button> is transpiled to React.createElement(Button, { active: true });

The same thing!

So, if you want to do a conditional prop, you can simply do something like that:

<Button active={this.state.view === 'default'}></Button>

You have a condition inside brackets. Means that, if your this.state.view is equal to default (true), active prop will be passwed down to the component with the true value. If not equal, with the false value.

Button Component, then, must someway handle this active prop. It can render the button element and, for instance, change it's style, pass the disabled prop... I created a fiddle to demonstrate this: https://jsfiddle.net/mrlew/5rsx40gu/

Answer 2

Actually, I guess it is a duplicated question that I answered it in this link sometimes ago. for this specific post, there is no need condition prop for a boolean value, because it works well like below:

const { booleanValue } = this.props;
return (
    <input checked={booleanValue} />
);

Or make your own boolean value:

const booleanValue = someThing === someOtherThing;
return (
    <input checked={booleanValue} />
);

Both of them work well, because when the booleanValue is false, react doesn't see the active prop, hence it does not exist, unless you pass checked prop to a specific component that the component will receive false checked prop from this.props.

But, if you wanna a have a prop on your specific component with a condition and if the condition returns false you don't want it there are two ways, I like second:

First:

<Component someProp={condition ? yourValue : undefined} />

Second:

<Component {...(condition && { someProp: yourValue })} />

Answer 3

In JSX, component properties (or props) compile to a plain JavaScript object. Prop names are used as the keys of the object, and prop values are stored under those keys. The first example from the JSX docs does a great good job demonstrating this. In your case {view === 'default' && true} is a raw value without an associated prop name. Without a prop name to use as a key (specified by the syntax name=), JSX has nowhere to put that value in the final props object.

However, JSX will accept props via your original syntax if the expression inside the curly braces evaluates to an object. For example, you could do {{ active: view === "default" }}. In this case JSX can get both the key and value that it needs from the provided object, so no active= is necessary.