ReactJs: Prevent multiple times button press

ReactJs: Prevent multiple times button press - javascript

In my React component I have a button meant to send some data over AJAX when clicked. I need to happen only the first time, i.e. to disable the button after its first use.
How I'm trying to do this:
var UploadArea = React.createClass({
getInitialState() {
return {
showUploadButton: true
};
},
disableUploadButton(callback) {
this.setState({ showUploadButton: false }, callback);
},
// This was simpler before I started trying everything I could think of
onClickUploadFile() {
if (!this.state.showUploadButton) {
return;
}
this.disableUploadButton(function() {
$.ajax({
[...]
});
});
},
render() {
var uploadButton;
if (this.state.showUploadButton) {
uploadButton = (
<button onClick={this.onClickUploadFile}>Send</button>
);
}
return (
<div>
{uploadButton}
</div>
);
}
});
What I think happens is the state variable showUploadButton not being updated right away, which the React docs says is expected.
How could I enforce the button to get disabled or go away altogether the instant it's being clicked?

The solution is to check the state immediately upon entry to the handler. React guarantees that setState inside interactive events (such as click) is flushed at browser event boundary. Ref: https://github.com/facebook/react/issues/11171#issuecomment-357945371
// In constructor
this.state = {
disabled : false
};
// Handler for on click
handleClick = (event) => {
if (this.state.disabled) {
return;
}
this.setState({disabled: true});
// Send
}
// In render
<button onClick={this.handleClick} disabled={this.state.disabled} ...>
{this.state.disabled ? 'Sending...' : 'Send'}
<button>

What you could do is make the button disabled after is clicked and leave it in the page (not clickable element).
To achieve this you have to add a ref to the button element
<button ref="btn" onClick={this.onClickUploadFile}>Send</button>
and then on the onClickUploadFile function disable the button
this.refs.btn.setAttribute("disabled", "disabled");
You can then style the disabled button accordingly to give some feedback to the user with
.btn:disabled{ /* styles go here */}
If needed make sure to reenable it with
this.refs.btn.removeAttribute("disabled");
Update: the preferred way of handling refs in React is with a function and not a string.
<button
ref={btn => { this.btn = btn; }}
onClick={this.onClickUploadFile}
>Send</button>
this.btn.setAttribute("disabled", "disabled");
this.btn.removeAttribute("disabled");
Update: Using react hooks
import {useRef} from 'react';
let btnRef = useRef();
const onBtnClick = e => {
if(btnRef.current){
btnRef.current.setAttribute("disabled", "disabled");
}
}
<button ref={btnRef} onClick={onBtnClick}>Send</button>
here is a small example using the code you provided
https://jsfiddle.net/69z2wepo/30824/

Tested as working one: http://codepen.io/zvona/pen/KVbVPQ
class UploadArea extends React.Component {
constructor(props) {
super(props)
this.state = {
isButtonDisabled: false
}
}
uploadFile() {
// first set the isButtonDisabled to true
this.setState({
isButtonDisabled: true
});
// then do your thing
}
render() {
return (
<button
type='submit'
onClick={() => this.uploadFile()}
disabled={this.state.isButtonDisabled}>
Upload
</button>
)
}
}
ReactDOM.render(<UploadArea />, document.body);

You can try using React Hooks to set the Component State.
import React, { useState } from 'react';
const Button = () => {
const [double, setDouble] = useState(false);
return (
<button
disabled={double}
onClick={() => {
// doSomething();
setDouble(true);
}}
/>
);
};
export default Button;
Make sure you are using ^16.7.0-alpha.x version or later of react and react-dom.
Hope this helps you!

If you want, just prevent to submit.
How about using lodash.js debounce
Grouping a sudden burst of events (like keystrokes) into a single one.
https://lodash.com/docs/4.17.11#debounce
<Button accessible={true}
onPress={_.debounce(async () => {
await this.props._selectUserTickets(this.props._accountId)
}, 1000)}
></Button>

If you disable the button during onClick, you basically get this. A clean way of doing this would be:
import React, { useState } from 'react';
import Button from '#material-ui/core/Button';
export default function CalmButton(props) {
const [executing, setExecuting] = useState(false);
const {
disabled,
onClick,
...otherProps
} = props;
const onRealClick = async (event) => {
setExecuting(true);
try {
await onClick();
} finally {
setExecuting(false);
}
};
return (
<Button
onClick={onRealClick}
disabled={executing || disabled}
{...otherProps}
/>
)
}
See it in action here: https://codesandbox.io/s/extended-button-that-disabled-itself-during-onclick-execution-mg6z8
We basically extend the Button component with the extra behaviour of being disabled during onClick execution. Steps to do this:
Create local state to capture if we are executing
Extract properties we tamper with (disabled, onClick)
Extend onClick operation with setting the execution state
Render the button with our overridden onClick, and extended disabled
NOTE: You should ensure that the original onClick operation is async aka it is returning a Promise.

By using event.target , you can disabled the clicked button.
Use arrow function when you create and call the function onClick. Don't forget to pass the event in parameter.
See my codePen
Here is the code:
class Buttons extends React.Component{
constructor(props){
super(props)
this.buttons = ['A','B','C','D']
}
disableOnclick = (e) =>{
e.target.disabled = true
}
render(){
return(
<div>
{this.buttons.map((btn,index) => (
<button type='button'
key={index}
onClick={(e)=>this.disableOnclick(e)}
>{btn}</button>
))}
</div>
)}
}
ReactDOM.render(<Buttons />, document.body);

const once = (f, g) => {
let done = false;
return (...args) => {
if (!done) {
done = true;
f(...args);
} else {
g(...args);
}
};
};
const exampleMethod = () => console.log("exampleMethod executed for the first time");
const errorMethod = () => console.log("exampleMethod can be executed only once")
let onlyOnce = once(exampleMethod, errorMethod);
onlyOnce();
onlyOnce();
output
exampleMethod executed for the first time
exampleMethod can be executed only once

You can get the element reference in the onClick callback and setAttribute from there, eg:
<Button
onClick={(e) => {
e.target.setAttribute("disabled", true);
this.handler();
}}
>
Submit
</Button>

Keep it simple and inline:
<button type="submit"
onClick={event => event.currentTarget.disabled = true}>
save
</button>
But! This will also disable the button, when the form calidation failed! So you will not be able to re-submit.
In this case a setter is better.
This fix this set the disabled in the onSubmit of the form:
// state variable if the form is currently submitting
const [submitting, setSubmitting] = useState(false);
// ...
return (
<form onSubmit={e => {
setSubmitting(true); // create a method to modify the element
}}>
<SubmitButton showLoading={submitting}>save</SubmitButton>
</form>
);
And the button would look like this:
import {ReactComponent as IconCog} from '../../img/icon/cog.svg';
import {useEffect, useRef} from "react";
export const SubmitButton = ({children, showLoading}) => {
const submitButton = useRef();
useEffect(() => {
if (showLoading) {
submitButton.current.disabled = true;
} else {
submitButton.current.removeAttribute("disabled");
}
}, [showLoading]);
return (
<button type="submit"
ref={submitButton}>
<main>
<span>{children}</span>
</main>
</button>
);
};

Another approach could be like so:
<button onClick={this.handleClick} disabled={isLoading ? "disabled" :""}>Send</button>

My approach is if event on processing do not execute anything.
class UploadArea extends React.Component {
constructor(props) {
super(props)
this.state = {
onProcess:false
}
}
uploadFile() {
if (!this.state.onProcess){
this.setState({
onProcess: true
});
// then do your thing
this.setState({
onProcess: false;
});
}
}
render() {
return (
<button
type='submit'
onClick={() => this.uploadFile()}>
Upload
</button>
)
}
}
ReactDOM.render(<UploadArea />, document.body);

Try with this code:
class Form extends React.Component {
constructor() {
this.state = {
disabled: false,
};
}
handleClick() {
this.setState({
disabled: true,
});
if (this.state.disabled) {
return;
}
setTimeout(() => this.setState({ disabled: false }), 2000);
}
render() {
return (
<button type="submit" onClick={() => this.handleClick()} disabled={this.state.disabled}>
Submit
</button>
);
}
}
ReactDOM.render(<Form />, document.getElementById('root'));

Related

onClick event is sending me to the correct method but from inside that method, it is not calling the inner component and it is lost

I have below EditAddressPage component which when I click on Manual Entry button, it should send me to the getAddressField method and from there go to <AddressFields /> component. I see when onCLick happens. it is sending me to the getAddressField method but from there it is lost. I was wondering if someone can give me a second though what is missing there because anything I've tried didn't work so far.
import AddressFields from 'components/widget/AddressFields';
import "./EditAddressPage.less";
const EditAddressPage = (props) => {
const getManualAddressEntry = () => {
return (
<AddressFields
{...props}
/>
)
}
return (
<div
<div className="body-container">
<span id="manually-enter-section">
{gt.gettext('Can’t find your address? ${0}',
<Button
color="primary"
iconSide="left"
iconSrc=""
onClick={() => getManualAddressEntry()}
size="medium"
variant="flat"
id="manual-entry-button"
>
Enter it manually
</Button>
)}
</span>
</div>
)
}
export default EditAddressPage;
export default class AddressFields extends PureRenderComponent {
constructor() {
super();
this.x = true;
this.y = undefined;
this.z = undefined;
}
componentWillMount() {
// something
}
componentDidMount() {
// something
}
componentWillUnmount() {
// something
}
render() {}
}

Your getManualAddressEntry() function does not render the component, it just returns it as a value.
To render it, you can use conditional rendering like so:
<Button
color="primary"
iconSide="left"
iconSrc=""
onClick={() => getManualAddressEntry()}
size="medium"
variant="flat"
id="manual-entry-button"
>
Enter it manually
</Button>
{buttonClicked && <AddressFields/>}
buttonClicked (or whatever name you prefer) should be a state that is initially set to false, and getManualAddressEntry() should toggle it between false and true
const [manualAddressOpen, setManualAddressOpen] = useState(false)
const getManualAddressEntry = () => {
setManualAddressOpen(!manualAddressOpen)
}
Example of conditional rendering:

Why am I getting an error here? React Todo list

I am trying to create a todo list using React but i cant seem to understand why I am getting the error: "Warning: Cannot update during an existing state transition (such as within render). Render methods should be a pure function of props and state."
Here's the code:
import React from 'react'
import ReactDOM from 'react-dom'
class Todo extends React.Component{
constructor(props){
super(props)
this.state = {
input: '',
list: []
}
this.handleChange = this.handleChange.bind(this)
this.reset = this.reset.bind(this)
this.removeItem = this.removeItem.bind(this)
this.add = this.add.bind(this)
}
add(){ //Adds a new task
const newItem = {
value: this.state.input,
id: Math.random + Math.random
};
const listed = [...this.state.list]
listed.push(newItem)
this.setState({
input: '',
list: listed
})
}
removeItem(id){ //deletes a task
const list = [...this.state.list]
const updatedList = list.filter(obj => {
return obj.id !== id
})
this.setState({
list: updatedList
})
}
handleChange(e){
this.setState({input: e.target.value})
}
reset(e){
e.preventDefault()
}
render(){
return (
<div>
<form action="" onSubmit={this.reset}>
<input type="text" value={this.state.input} placeholder='Enter a task..' onChange={this.handleChange} />
<button onClick={this.add}>Add Task</button>
{this.state.list.map(item => { //updates when a task is added or removed
return (
<div key={item.id}>
<h1>{item.value}</h1>
<button onClick={this.removeItem(item.id)}>X</button>
</div>
)
})}
</form>
</div>
)
}
}
ReactDOM.render(<Todo />,document.getElementById('root'))

Because you are calling removeItem on render. It needs to be wrapped in a separate function:
<button onClick={() => this.removeItem(item.id)}>X</button>
So that you only call it onClick and not on render.

<button onClick={this.removeItem(item.id)}>X</button>
In this button the event handler you have provided runs immediately due to the presents of the () at the end. To prevent this and still provide your argument item.id you can enclose the handler this.removeItem(item.id) with in another function.
I like the arrow function for this so mine looks like this
<button onClick={ ()=>this.removeItem(item.id) }>X</button>.
Math.random + Math.random is not returning a number like you would want for the element key. This is because your have neglected to include () at telling JS to run the function and return an int.
After making these changes, I ran it in codepen.io and it seemed to work fine.

How can I assign an integer value to a button using React and JSX?

So basically I am practicing React and trying to make simple components communicate with the parent component. I have come to an issue where I can't figure out how to change the value of a button into an integer so I can make simple calulations. I have tried setting the value with parseInt and without it as well, but I can't seen to understand, pressing the "add" button gives me a NaN. Could anyone help me?
This is what I have in my App.js
import './App.css';
import Input from "./Components/Input"
class App extends React.Component {
constructor(props){
super(props);
this.state = {
number: 10
}
}
number1 = (event) => {
this.setState({number: event.target.value})
}
number2 = (event) => {
this.setState({number: event.target.value})
if (this.state.number < 1){
this.setState({number: 0})
}
}
add = (event) => {
this.setState({number: this.number+1})
}
render() {
return (
<div>
<Input number={this.state.number}
number1={this.number1}
number2={this.number2}
add={this.add} />
</div>
);
}
}
export default App;
and this is what I have in Input.js
let numberValue1= parseInt("8")
let numberValue2= parseInt("17")
const Input = ({number, number1, number2, add}) =>{
return(
<div>
<h1>{number}</h1>
<button value={numberValue1} onClick={number1}>Add</button>
<button value={numberValue2} onClick={number2}>Subtract</button>
<button onClick={add}>Subtract</button>
</div>
);
}
export default Input;```

What you're trying to do is fundamentally a bit strange - you probably shouldn't be assigning values to button elements, as it's doesn't really make sense for them to have one (if I were to ask you "what is this button's value?" what would you think I meant?).
Instead, it makes more sense to pass a function to the button that accepts an argument.
For example:
const ParentComponent = ()=> {
const [number, setNumber] = useState(0);
const add = (amount)=> {
setNumber(number + amount);
};
return (
<MyInput number={number} onAddButtonClicked={add} />
)
}
const MyInput = ({number, onAddButtonClicked}) => {
<div>
<h1>{number}</h1>
<button onClick={ ()=> onAddButtonClicked(1) }>Add</button>
<button onClick={ ()=> onAddButtonClicked(-1) }>Subtract</button>
</div>
}

ReactJS - Stop button onClick from taking focus from <input>

So I've set up a file to setState() for onBlur and onFocus in the SocialPost.js file. But when I onClick a <div> in the SocialPostList.js (the parent) where it activates the parameterClicked() function in the SocialPost.js file, the <input> in SocialPost.js becomes blurred.
How do I make it so that the <button> onClick in SocialPostList.js does not take the focus() from the <input> in SocialPost.js?
I've tried e.preventDefault() and e.stopPropagation() without success. The files are below, any help would be appreciated!!!
SocialPostList.js
import React, { Component } from 'react'
import { graphql, gql } from 'react-apollo'
import SocialPost from './SocialPost'
class SocialPostList extends Component {
render() {
const PostListArray = () => {
return(
<div onClick={(e) => {e.preventDefault(); e.stopPropagation()}}>
{this.props.allParametersQuery.allParameters.map((parameter, index) => (
<div
key={index}
onClick={(e) => {e.preventDefault();e.stopPropagation();this.child.parameterClicked(parameter.param, parameter.id)}}
>{'{{' + parameter.param + '}}'}</div>
))}
</div>)
}
return (
<div>
...
<PostListArray />
{this.props.allSocialPostsQuery.allSocialPosts.map((socialPost, index) => (
<SocialPost
ref={instance => {this.child = instance}}
key={socialPost.id}
socialPost={socialPost}
index={index}
deleteSocialPost={this._handleDeleteSocialPost}
updateSocialPost={this._handleUpdateSocialPost}
allParametersQuery={this.props.allParametersQuery}/>
))}
...
</div>
)
}
}
const ALL_SOCIAL_POSTS_QUERY = gql`
query AllSocialPostsQuery {
allSocialPosts {
id
default
message
}}`
export default graphql(ALL_SOCIAL_POSTS_QUERY, {name: 'allSocialPostsQuery'})(SocialPostList)
SocialPost.js
import React, { Component } from 'react'
class SocialPost extends Component {
constructor(props) {
super(props)
this.state = {
message: this.props.socialPost.message,
focus: false
}
this._onBlur = this._onBlur.bind(this)
this._onFocus = this._onFocus.bind(this)
}
_onBlur() {
setTimeout(() => {
if (this.state.focus) {
this.setState({ focus: false });
}}, 0);
}
_onFocus() {
if (!this.state.focus) {
this.setState({ focus: true });
}
}
render() {
return (
<div className='socialpostbox mb1'>
<div className='flex'>
<input
onFocus={this._onFocus}
onBlur={this._onBlur}
type='text'
value={this.state.message}
onChange={(e) => { this.setState({ message: e.target.value})}}/>
</div>
</div>
)
}
parameterClicked = (parameterParam) =>{
if (!this.state.focus) return
let message = this.state.message
let newMessage = message.concat(' ' + parameterParam)
this.setState({ message: newMessage })
}
export default SocialPost

Well, I don't think that's a React thing. It appears the blur event fires before the onClick, so the latter cannot prevent the former, and I'd expect event.stopPropagation to stop events bubbling from child to parent, not the other way around. In other words, I don't know how to stop it.
In all fairness this behaviour is expected - clicking somewhere else makes you lose focus. That said, here and elsewhere a solution is presented where you set up a flag on mouse down. Then, when blur fires, if it encounters the 'click flag' it may abstain from producing effects and may even refocus back.
If you choose to refocus, it is trivial to save a reference to the button or input, or querySelecting it (it's not too late or anything like that). Just be cautious that it is all too easy to set focus traps or mess up navigation for screen readers when you mix focus and javascript.

Show or hide element in React

I am messing around with React.js for the first time and cannot find a way to show or hide something on a page via click event. I am not loading any other library to the page, so I am looking for some native way using the React library. This is what I have so far. I would like to show the results div when the click event fires.
var Search= React.createClass({
handleClick: function (event) {
console.log(this.prop);
},
render: function () {
return (
<div className="date-range">
<input type="submit" value="Search" onClick={this.handleClick} />
</div>
);
}
});
var Results = React.createClass({
render: function () {
return (
<div id="results" className="search-results">
Some Results
</div>
);
}
});
React.renderComponent(<Search /> , document.body);

React circa 2020
In the onClick callback, call the state hook's setter function to update the state and re-render:
const Search = () => {
const [showResults, setShowResults] = React.useState(false)
const onClick = () => setShowResults(true)
return (
<div>
<input type="submit" value="Search" onClick={onClick} />
{ showResults ? <Results /> : null }
</div>
)
}
const Results = () => (
<div id="results" className="search-results">
Some Results
</div>
)
ReactDOM.render(<Search />, document.querySelector("#container"))
<script src="https://cdnjs.cloudflare.com/ajax/libs/react/16.13.1/umd/react.production.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/react-dom/16.13.1/umd/react-dom.production.min.js"></script>
<div id="container">
<!-- This element's contents will be replaced with your component. -->
</div>
JSFiddle
React circa 2014
The key is to update the state of the component in the click handler using setState. When the state changes get applied, the render method gets called again with the new state:
var Search = React.createClass({
getInitialState: function() {
return { showResults: false };
},
onClick: function() {
this.setState({ showResults: true });
},
render: function() {
return (
<div>
<input type="submit" value="Search" onClick={this.onClick} />
{ this.state.showResults ? <Results /> : null }
</div>
);
}
});
var Results = React.createClass({
render: function() {
return (
<div id="results" className="search-results">
Some Results
</div>
);
}
});
ReactDOM.render( <Search /> , document.getElementById('container'));
<script src="https://cdnjs.cloudflare.com/ajax/libs/react/15.6.2/react.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/react-dom/15.6.2/react-dom.min.js"></script>
<div id="container">
<!-- This element's contents will be replaced with your component. -->
</div>
JSFiddle

<style type="text/css">
.hidden { display:none; }
</style>
const Example = props =>
<div className={props.shouldHide? 'hidden' : undefined}>Hello</div>

Here is an alternative syntax for the ternary operator:
{ this.state.showMyComponent ? <MyComponent /> : null }
is equivalent to:
{ this.state.showMyComponent && <MyComponent /> }
Learn why
Also alternative syntax with display: 'none';
<MyComponent style={this.state.showMyComponent ? {} : { display: 'none' }} />
However, if you overuse display: 'none', this leads to DOM pollution and ultimately slows down your application.

Here is my approach.
import React, { useState } from 'react';
function ToggleBox({ title, children }) {
const [isOpened, setIsOpened] = useState(false);
function toggle() {
setIsOpened(wasOpened => !wasOpened);
}
return (
<div className="box">
<div className="boxTitle" onClick={toggle}>
{title}
</div>
{isOpened && (
<div className="boxContent">
{children}
</div>
)}
</div>
);
}
In code above, to achieve this, I'm using code like:
{opened && <SomeElement />}
That will render SomeElement only if opened is true. It works because of the way how JavaScript resolve logical conditions:
true && true && 2; // will output 2
true && false && 2; // will output false
true && 'some string'; // will output 'some string'
opened && <SomeElement />; // will output SomeElement if `opened` is true, will output false otherwise (and false will be ignored by react during rendering)
// be careful with 'falsy' values eg
const someValue = [];
someValue.length && <SomeElement /> // will output 0, which will be rednered by react
// it'll be better to:
someValue.length > 0 && <SomeElement /> // will render nothing as we cast the value to boolean
Reasons for using this approach instead of CSS 'display: none';
While it might be 'cheaper' to hide an element with CSS - in such case 'hidden' element is still 'alive' in react world (which might make it actually way more expensive)
it means that if props of the parent element (eg. <TabView>) will change - even if you see only one tab, all 5 tabs will get re-rendered
the hidden element might still have some lifecycle methods running - eg. it might fetch some data from the server after every update even tho it's not visible
the hidden element might crash the app if it'll receive incorrect data. It might happen as you can 'forget' about invisible nodes when updating the state
you might by mistake set wrong 'display' style when making element visible - eg. some div is 'display: flex' by default, but you'll set 'display: block' by mistake with display: invisible ? 'block' : 'none' which might break the layout
using someBoolean && <SomeNode /> is very simple to understand and reason about, especially if your logic related to displaying something or not gets complex
in many cases, you want to 'reset' element state when it re-appears. eg. you might have a slider that you want to set to initial position every time it's shown. (if that's desired behavior to keep previous element state, even if it's hidden, which IMO is rare - I'd indeed consider using CSS if remembering this state in a different way would be complicated)

with the newest version react 0.11 you can also just return null to have no content rendered.
Rendering to null

This is a nice way to make use of the virtual DOM:
class Toggle extends React.Component {
state = {
show: true,
}
toggle = () => this.setState((currentState) => ({show: !currentState.show}));
render() {
return (
<div>
<button onClick={this.toggle}>
toggle: {this.state.show ? 'show' : 'hide'}
</button>
{this.state.show && <div>Hi there</div>}
</div>
);
}
}
Example here
Using React hooks:
const Toggle = () => {
const [show, toggleShow] = React.useState(true);
return (
<div>
<button
onClick={() => toggleShow(!show)}
>
toggle: {show ? 'show' : 'hide'}
</button>
{show && <div>Hi there</div>}
</div>
)
}
Example here

I created a small component that handles this for you: react-toggle-display
It sets the style attribute to display: none !important based on the hide or show props.
Example usage:
var ToggleDisplay = require('react-toggle-display');
var Search = React.createClass({
getInitialState: function() {
return { showResults: false };
},
onClick: function() {
this.setState({ showResults: true });
},
render: function() {
return (
<div>
<input type="submit" value="Search" onClick={this.onClick} />
<ToggleDisplay show={this.state.showResults}>
<Results />
</ToggleDisplay>
</div>
);
}
});
var Results = React.createClass({
render: function() {
return (
<div id="results" className="search-results">
Some Results
</div>
);
}
});
React.renderComponent(<Search />, document.body);

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.

You set a boolean value in the state (e.g. 'show)', and then do:
var style = {};
if (!this.state.show) {
style.display = 'none'
}
return <div style={style}>...</div>

A simple method to show/hide elements in React using Hooks
const [showText, setShowText] = useState(false);
Now, let's add some logic to our render method:
{showText && <div>This text will show!</div>}
And
onClick={() => setShowText(!showText)}
Good job.

I was able to use css property "hidden". Don't know about possible drawbacks.
export default function App() {
const [hidden, setHidden] = useState(false);
return (
<div>
<button onClick={() => setHidden(!hidden)}>HIDE</button>
<div hidden={hidden}>hidden component</div>
</div>
);
}

Best practice is below according to the documentation:
{this.state.showFooter && <Footer />}
Render the element only when the state is valid.

Simple hide/show example with React Hooks: (srry about no fiddle)
const Example = () => {
const [show, setShow] = useState(false);
return (
<div>
<p>Show state: {show}</p>
{show ? (
<p>You can see me!</p>
) : null}
<button onClick={() => setShow(!show)}>
</div>
);
};
export default Example;

class FormPage extends React.Component{
constructor(props){
super(props);
this.state = {
hidediv: false
}
}
handleClick = (){
this.setState({
hidediv: true
});
}
render(){
return(
<div>
<div className="date-range" hidden = {this.state.hidediv}>
<input type="submit" value="Search" onClick={this.handleClick} />
</div>
<div id="results" className="search-results" hidden = {!this.state.hidediv}>
Some Results
</div>
</div>
);
}
}

I start with this statement from the React team:
In React, you can create distinct components that encapsulate behaviour
you need. Then, you can render only some of them, depending on the
state of your application.
Conditional rendering in React works the same way conditions work in
JavaScript. Use JavaScript operators like if or the conditional
operator to create elements representing the current state, and let
React update the UI to match them.
You basically need to show the component when the button gets clicked, you can do it two ways, using pure React or using CSS, using pure React way, you can do something like below code in your case, so in the first run, results are not showing as hideResults is true, but by clicking on the button, state gonna change and hideResults is false and the component get rendered again with the new value conditions, this is very common use of changing component view in React...
var Search = React.createClass({
getInitialState: function() {
return { hideResults: true };
},
handleClick: function() {
this.setState({ hideResults: false });
},
render: function() {
return (
<div>
<input type="submit" value="Search" onClick={this.handleClick} />
{ !this.state.hideResults && <Results /> }
</div> );
}
});
var Results = React.createClass({
render: function() {
return (
<div id="results" className="search-results">
Some Results
</div>);
}
});
ReactDOM.render(<Search />, document.body);
If you want to do further study in conditional rendering in React, have a look here.

class Toggle extends React.Component {
state = {
show: true,
}
render() {
const {show} = this.state;
return (
<div>
<button onClick={()=> this.setState({show: !show })}>
toggle: {show ? 'show' : 'hide'}
</button>
{show && <div>Hi there</div>}
</div>
);
}
}

If you would like to see how to TOGGLE the display of a component checkout this fiddle.
http://jsfiddle.net/mnoster/kb3gN/16387/
var Search = React.createClass({
getInitialState: function() {
return {
shouldHide:false
};
},
onClick: function() {
console.log("onclick");
if(!this.state.shouldHide){
this.setState({
shouldHide: true
})
}else{
this.setState({
shouldHide: false
})
}
},
render: function() {
return (
<div>
<button onClick={this.onClick}>click me</button>
<p className={this.state.shouldHide ? 'hidden' : ''} >yoyoyoyoyo</p>
</div>
);
}
});
ReactDOM.render( <Search /> , document.getElementById('container'));

Use ref and manipulate CSS
One way could be to use React's ref and manipulate CSS class using the browser's API. Its benefit is to avoid rerendering in React if the sole purpose is to hide/show some DOM element on the click of a button.
// Parent.jsx
import React, { Component } from 'react'
export default class Parent extends Component {
constructor () {
this.childContainer = React.createRef()
}
toggleChild = () => {
this.childContainer.current.classList.toggle('hidden')
}
render () {
return (
...
<button onClick={this.toggleChild}>Toggle Child</button>
<div ref={this.childContainer}>
<SomeChildComponent/>
</div>
...
);
}
}
// styles.css
.hidden {
display: none;
}
PS Correct me if I am wrong. :)

In some cases higher order component might be useful:
Create higher order component:
export var HidableComponent = (ComposedComponent) => class extends React.Component {
render() {
if ((this.props.shouldHide!=null && this.props.shouldHide()) || this.props.hidden)
return null;
return <ComposedComponent {...this.props} />;
}
};
Extend your own component:
export const MyComp= HidableComponent(MyCompBasic);
Then you can use it like this:
<MyComp hidden={true} ... />
<MyComp shouldHide={this.props.useSomeFunctionHere} ... />
This reduces a bit boilerplate and enforces sticking to naming conventions, however please be aware of that MyComp will still be instantiated - the way to omit is was mentioned earlier:
{ !hidden && <MyComp ... /> }

If you use bootstrap 4, you can hide element that way
className={this.state.hideElement ? "invisible" : "visible"}

Use rc-if-else module
npm install --save rc-if-else
import React from 'react';
import { If } from 'rc-if-else';
class App extends React.Component {
render() {
return (
<If condition={this.props.showResult}>
Some Results
</If>
);
}
}

Use this lean and short syntax:
{ this.state.show && <MyCustomComponent /> }

Here comes the simple, effective and best solution with a Classless React Component for show/hide the elements. Use of React-Hooks which is available in the latest create-react-app project that uses React 16
import React, {useState} from 'react';
function RenderPara(){
const [showDetail,setShowDetail] = useState(false);
const handleToggle = () => setShowDetail(!showDetail);
return (
<React.Fragment>
<h3>
Hiding some stuffs
</h3>
<button onClick={handleToggle}>Toggle View</button>
{showDetail && <p>
There are lot of other stuffs too
</p>}
</React.Fragment>)
}
export default RenderPara;
Happy Coding :)

//use ternary condition
{ this.state.yourState ? <MyComponent /> : null }
{ this.state.yourState && <MyComponent /> }
{ this.state.yourState == 'string' ? <MyComponent /> : ''}
{ this.state.yourState == 'string' && <MyComponent /> }
//Normal condition
if(this.state.yourState){
return <MyComponent />
}else{
return null;
}
<button onClick={()=>this.setState({yourState: !this.props.yourState}>Toggle View</button>

Just figure out a new and magic way with using(useReducer) for functional components
const [state, handleChangeState] = useReducer((state) => !state, false);
change state

This can also be achieved like this (very easy way)
class app extends Component {
state = {
show: false
};
toggle= () => {
var res = this.state.show;
this.setState({ show: !res });
};
render() {
return(
<button onClick={ this.toggle }> Toggle </button>
{
this.state.show ? (<div> HELLO </div>) : null
}
);
}

this example shows how you can switch between components by using a toggle which switches after every 1sec
import React ,{Fragment,Component} from "react";
import ReactDOM from "react-dom";
import "./styles.css";
const Component1 = () =>(
<div>
<img
src="https://i.pinimg.com/originals/58/df/1d/58df1d8bf372ade04781b8d4b2549ee6.jpg" />
</div>
)
const Component2 = () => {
return (
<div>
<img
src="http://www.chinabuddhismencyclopedia.com/en/images/thumb/2/2e/12ccse.jpg/250px-
12ccse.jpg" />
</div>
)
}
class App extends Component {
constructor(props) {
super(props);
this.state = {
toggleFlag:false
}
}
timer=()=> {
this.setState({toggleFlag:!this.state.toggleFlag})
}
componentDidMount() {
setInterval(this.timer, 1000);
}
render(){
let { toggleFlag} = this.state
return (
<Fragment>
{toggleFlag ? <Component1 /> : <Component2 />}
</Fragment>
)
}
}
const rootElement = document.getElementById("root");
ReactDOM.render(<App />, rootElement);

The application of states and effects has and must be encapsulated in the same component, for this reason, there is nothing better than creating a custom component as a hook to solve in this case whether to make particular blocks or elements visible or invisible.
// hooks/useOnScreen.js
import { useState, useEffect } from "react"
const useOnScreen = (ref, rootMargin = "0px") => {
const [isVisible, setIsVisible] = useState(false)
useEffect(() => {
const observer = new IntersectionObserver(
([entry]) => {
setIsVisible(entry.isIntersecting)
},
{
rootMargin
}
);
const currentElement = ref?.current
if (currentElement) {
observer.observe(currentElement)
}
return () => {
observer.unobserve(currentElement)
}
}, [])
return isVisible
}
export default useOnScreen
Then the custom hook is embedded inside the component
import React, { useRef } from "react";
import useOnScreen from "hooks/useOnScreen";
const MyPage = () => {
const ref = useRef(null)
const isVisible = useOnScreen(ref)
const onClick = () => {
console.log("isVisible", isVisible)
}
return (
<div ref={ref}>
<p isVisible={isVisible}>
Something is visible
</p>
<a
href="#"
onClick={(e) => {
e.preventDefault();
onClick(onClick)
}}
>
Review
</a>
</div>
)
}
export default MyPage
The ref variable, controlled by the useRef hook, allows us to capture the location in the DOM of the block that we want to control, then the isVisible variable, controlled by the useOnScreen hook, allows us to make the inside the block I signal by the useRef hook.
I believe that this implementation of the useState, useEfect, and useRef hooks allows you to avoid component rendering by separating them using custom hooks.
Hoping that this knowledge will be of use to you.

It is very simple to hide and show the elements in react.
There are multiple ways but I will show you two.
Way 1:
const [isVisible, setVisible] = useState(false)
let onHideShowClick = () =>{
setVisible(!isVisible)
}
return (<div>
<Button onClick={onHideShowClick} >Hide/Show</Button>
{(isVisible) ? <p>Hello World</p> : ""}
</div>)
Way 2:
const [isVisible, setVisible] = useState(false)
let onHideShowClick = () =>{
setVisible(!isVisible)
}
return (<div>
<Button onClick={onHideShowClick} >Hide/Show</Button>
<p style={{display: (isVisible) ? 'block' : 'none'}}>Hello World</p>
</div>)
It is just working like if and else.
In Way one, it will remove and re-render elements in Dom.
In the Second way you are just displaying elements as false or true.
Thank you.

You've to do the small change in the code for continuously hiding and showing
const onClick = () => {setShowResults(!showResults}
Problem will be solved
const Search = () => {
const [showResults, setShowResults] = React.useState(false)
const onClick = () => setShowResults(true)
const onClick = () => {setShowResults(!showResults}
return (
<div>
<input type="submit" value="Search" onClick={onClick} />
{ showResults ? <Results /> : null }
</div>
)
}
const Results = () => (
<div id="results" className="search-results">
Some Results
</div>
)
ReactDOM.render(<Search />, document.querySelector("#container"))
<script src="https://cdnjs.cloudflare.com/ajax/libs/react/16.13.1/umd/react.production.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/react-dom/16.13.1/umd/react-dom.production.min.js"></script>
<div id="container">
<!-- This element's contents will be replaced with your component. -->
</div>
```

We Keep Coding

JavaScript is the programming language of the Web.

ReactJs: Prevent multiple times button press - javascript

You can get the element reference in the onClick callback and setAttribute from there, eg: <Button onClick={(e) => { e.target.setAttribute("disabled", true); this.handler(); }} > Submit </Button>

Another approach could be like so: <button onClick={this.handleClick} disabled={isLoading ? "disabled" :""}>Send</button>

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