There are several great answers already, but I don't think they've been explained very well and several of the methods given contain some gotchas that might trip people up. So I'm going to go over the three main ways (plus one off-topic option) to do this and explain the pros and cons. I'm mostly writing this because Option 1 was recommended a lot and there's a lot of potential issues with that option if not used correctly.
Option 1: Conditional Rendering in the parent.
I don't like this method unless you're only going to render the component one time and leave it there. The issue is it will cause react to create the component from scratch every time you toggle the visibility.
Here's the example. LogoutButton or LoginButton are being conditionally rendered in the parent LoginControl. If you run this you'll notice the constructor is getting called on each button click. https://codepen.io/Kelnor/pen/LzPdpN?editors=1111
class LoginControl extends React.Component {
constructor(props) {
super(props);
this.handleLoginClick = this.handleLoginClick.bind(this);
this.handleLogoutClick = this.handleLogoutClick.bind(this);
this.state = {isLoggedIn: false};
}
handleLoginClick() {
this.setState({isLoggedIn: true});
}
handleLogoutClick() {
this.setState({isLoggedIn: false});
}
render() {
const isLoggedIn = this.state.isLoggedIn;
let button = null;
if (isLoggedIn) {
button = <LogoutButton onClick={this.handleLogoutClick} />;
} else {
button = <LoginButton onClick={this.handleLoginClick} />;
}
return (
<div>
<Greeting isLoggedIn={isLoggedIn} />
{button}
</div>
);
}
}
class LogoutButton extends React.Component{
constructor(props, context){
super(props, context)
console.log('created logout button');
}
render(){
return (
<button onClick={this.props.onClick}>
Logout
</button>
);
}
}
class LoginButton extends React.Component{
constructor(props, context){
super(props, context)
console.log('created login button');
}
render(){
return (
<button onClick={this.props.onClick}>
Login
</button>
);
}
}
function UserGreeting(props) {
return <h1>Welcome back!</h1>;
}
function GuestGreeting(props) {
return <h1>Please sign up.</h1>;
}
function Greeting(props) {
const isLoggedIn = props.isLoggedIn;
if (isLoggedIn) {
return <UserGreeting />;
}
return <GuestGreeting />;
}
ReactDOM.render(
<LoginControl />,
document.getElementById('root')
);
Now React is pretty quick at creating components from scratch. However, it still has to call your code when creating it. So if your constructor, componentDidMount, render, etc code is expensive, then it'll significantly slow down showing the component. It also means you cannot use this with stateful components where you want the state to be preserved when hidden (and restored when displayed.) The one advantage is that the hidden component isn't created at all until it's selected. So hidden components won't delay your initial page load. There may also be cases where you WANT a stateful component to reset when toggled. In which case this is your best option.
Option 2: Conditional Rendering in the child
This creates both components once. Then short circuits the rest of the render code if the component is hidden. You can also short circuit other logic in other methods using the visible prop. Notice the console.log in the codepen page. https://codepen.io/Kelnor/pen/YrKaWZ?editors=0011
class LoginControl extends React.Component {
constructor(props) {
super(props);
this.handleLoginClick = this.handleLoginClick.bind(this);
this.handleLogoutClick = this.handleLogoutClick.bind(this);
this.state = {isLoggedIn: false};
}
handleLoginClick() {
this.setState({isLoggedIn: true});
}
handleLogoutClick() {
this.setState({isLoggedIn: false});
}
render() {
const isLoggedIn = this.state.isLoggedIn;
return (
<div>
<Greeting isLoggedIn={isLoggedIn} />
<LoginButton isLoggedIn={isLoggedIn} onClick={this.handleLoginClick}/>
<LogoutButton isLoggedIn={isLoggedIn} onClick={this.handleLogoutClick}/>
</div>
);
}
}
class LogoutButton extends React.Component{
constructor(props, context){
super(props, context)
console.log('created logout button');
}
render(){
if(!this.props.isLoggedIn){
return null;
}
return (
<button onClick={this.props.onClick}>
Logout
</button>
);
}
}
class LoginButton extends React.Component{
constructor(props, context){
super(props, context)
console.log('created login button');
}
render(){
if(this.props.isLoggedIn){
return null;
}
return (
<button onClick={this.props.onClick}>
Login
</button>
);
}
}
function UserGreeting(props) {
return <h1>Welcome back!</h1>;
}
function GuestGreeting(props) {
return <h1>Please sign up.</h1>;
}
function Greeting(props) {
const isLoggedIn = props.isLoggedIn;
if (isLoggedIn) {
return <UserGreeting />;
}
return <GuestGreeting />;
}
ReactDOM.render(
<LoginControl />,
document.getElementById('root')
);
Now, if the initialization logic is quick and the children are stateless, then you won't see a difference in performance or functionality. However, why make React create a brand new component every toggle anyway? If the initialization is expensive however, Option 1 will run it every time you toggle a component which will slow the page down when switching. Option 2 will run all of the component's inits on first page load. Slowing down that first load. Should note again. If you're just showing the component one time based on a condition and not toggling it, or you want it to reset when toggledm, then Option 1 is fine and probably the best option.
If slow page load is a problem however, it means you've got expensive code in a lifecycle method and that's generally not a good idea. You can, and probably should, solve the slow page load by moving the expensive code out of the lifecycle methods. Move it to an async function that's kicked off by ComponentDidMount and have the callback put it in a state variable with setState(). If the state variable is null and the component is visible then have the render function return a placeholder. Otherwise render the data. That way the page will load quickly and populate the tabs as they load. You can also move the logic into the parent and push the results to the children as props. That way you can prioritize which tabs get loaded first. Or cache the results and only run the logic the first time a component is shown.
Option 3: Class Hiding
Class hiding is probably the easiest to implement. As mentioned you just create a CSS class with display: none and assign the class based on prop. The downside is the entire code of every hidden component is called and all hidden components are attached to the DOM. (Option 1 doesn't create the hidden components at all. And Option 2 short circuits unnecessary code when the component is hidden and removes the component from the DOM completely.) It appears this is faster at toggling visibility according some tests done by commenters on other answers but I can't speak to that.
Option 4: One component but change Props. Or maybe no component at all and cache HTML.
This one won't work for every application and it's off topic because it's not about hiding components, but it might be a better solution for some use cases than hiding. Let's say you have tabs. It might be possible to write one React Component and just use the props to change what's displayed in the tab. You could also save the JSX to state variables and use a prop to decide which JSX to return in the render function. If the JSX has to be generated then do it and cache it in the parent and send the correct one as a prop. Or generate in the child and cache it in the child's state and use props to select the active one.
