Develop Reference

Passing React State

I have a modal that I want to appear on a button click. This works fine, however, in order for it to be positioned correctly, I need to move the modal component to be rendered in a parent component.
I'm unsure how to do this as I'm not sure how I would pass the state down to the component where the button click is created. I'm also unsure of whether this is about passing a state or using a button onClick outside the component the state is defined in.
My code works and does what I want, but, I need <DeleteWarehouseModal show={show} /> to be in the parent component below. How can I do this?
The parent component where I want to render my modal:
function WarehouseComponent(props) {
return (
<section>
<div>
<WarehouseListItems warehouseData={props.warehouseData} />
//Modal component should go here
</div>
</section>
);
}
The component (WarehouseListItems) where the modal is currently being rendered:
function WarehouseListItem(props) {
const [show, setShow] = useState(false);
return (
<>
//some code necessary to the project, but, irrelevant to this issue
<Link>
<div onClick={() => setShow(true)}></div>
</Link>
<DeleteWarehouseModal show={show} />
</>
);
}
The modal:
const DeleteWarehouseModal = (props) => {
if (!props.show) {
return null;
}
return (
//some elements here
);
};

Yes, you can move the state and it's handler in WarehouseComponent component and pass the handler down to child component, which can change the state in parent component.
WarehouseComponent :-
function WarehouseComponent(props) {
const [show, setShow] = useState(false);
const toggleShow = () => {
setShow(state => !state);
}
return (
<section>
<div>
<WarehouseListItems
warehouseData={props.warehouseData}
toggleShow={toggleShow}
/>
<DeleteWarehouseModal show={show} />
</div>
</section>
);
}
WarehouseListItems : -
function WarehouseListItem(props) {
const {toggleShow} = props;
return (
<>
//some code necessary to the project, but, irrelevant to this issue
<Link>
<div onClick={() => toggleShow()}></div>
</Link>
</>
);
}

Modal component does not render on a custom button component

I am trying to render a custom and dynamic modal on button clicks. For example, when a "Game" button is clicked, I would like a modal to render with specfics about the game and when a "Bank" button is clicked, I would like the modal to populate with specfics about a bank.
First, when I add an onClick function to a custom button component, the modal does not render. However, when I put the onClick function on a regular button, the modal does render. How can I simply add an onClick function on any component to render a dynamic modal?
Second, I would like to populate each modal with differnet data. For example, a "Game" button would populate the modal with a title of "Game" and so on. I'm using props to do this, but is that the best solution?
Here is the code I have so far, but it is broken when I add the onClick function to components.
// Navbar.js
import { ModalContext } from '../contexts/ModalContext'
function Navbar() {
const [showModal, updateShowModal] = React.useState(false)
const toggleModal = () => updateShowModal((state) => !state)
return(
<ModalContext.Provider value={{ showModal, toggleModal }}>
<Modal
title="Title"
canShow={showModal}
updateModalState={toggleModal}
/>
</ModalContext.Provider>
)
// does not render a modal
<Button
onClick={toggleModal}
type="navItem"
label="Game"
icon="windows"
/>
// render a modal
<button onClick={toggleModal}>Show Modal</button>
)
}
import { ModalContext } from '../contexts/ModalContext'
// Modal.js
const Modal = ({ title }) => {
return (
<ModalContext.Consumer>
{(context) => {
if (context.showModal) {
return (
<div style={modalStyles}>
<h1>{title}</h1>
<button onClick={context.toggleModal}>X</button>
</div>
)
}
return null
}}
</ModalContext.Consumer>
)
}
// modalContext.js
export const ModalContext = React.createContext()
// Button.js
function Button({ label, type = 'default', icon }) {
return (
<ButtonStyle buttonType={type}>
{setIcon(icon)}
{label}
</ButtonStyle>
)
}

First problem:
I think the onClick prop of the <Button> component is not pointing to the onClick of the actual HTML button inside the component.
Could you please check that? And if you think It's been set up in the right way, then can you share the code of the component?
Second Problem
Yes, there's another way to do that. And I think it's React Composition. You can build the modal as the following:
<Modal
showModal={showModal}
updateModalState={toggleModal}
>
<div className="modal__header">{title}</div>
<div className="modal__body">{body}</div>
<div className="modal__footer">{footer}</div>
</Modal>
I think this pattern will give you more control over that component.

Issue
You are not passing the onClick prop through to the styled button component.
Solution
Given style-component button:
const ButtonStyle = styled.button``;
The custom Button component needs to pass all button props on to the ButtonStyle component.
// Button.js
function Button({ label, type='default', icon, onClick }) {
return (
<ButtonStyle buttonType={type} onClick={onClick}>
{setIcon(icon)}
{label}
</ButtonStyle>
)
}
If there are other button props then you can use the Spread syntax to collect them into a single object that can then be spread into the ButtonStyle component.
// Button.js
function Button({ label, type = 'default', icon, ...props }) {
return (
<ButtonStyle buttonType={type} {...props}>
{setIcon(icon)}
{label}
</ButtonStyle>
)
}
Second Question
For the second issue I suggest encapsulating the open/close/title state entirely in the modal context provider, along with the Modal component.
Here's an example implementation:
const ModalContext = React.createContext({
openModal: () => {},
});
const Modal = ({ title, onClose}) => (
<>
<h1>{title}</h1>
<button onClick={onClose}>X</button>
</>
)
const ModalProvider = ({ children }) => {
const [showModal, setShowModal] = React.useState(false);
const [title, setTitle] = React.useState('');
const openModal = (title) => {
setShowModal(true);
setTitle(title);
}
const closeModal = () => setShowModal(false);
return (
<ModalContext.Provider value={{ openModal }}>
{children}
{showModal && <Modal title={title} onClose={closeModal} />}
</ModalContext.Provider>
)
}
Example consumer to set/open a modal:
const OpenModalButton = ({ children }) => {
const { openModal } = useContext(ModalContext);
return <button onClick={() => openModal(children)}>{children}</button>
}
Example usage:
function App() {
return (
<ModalProvider>
<div className="App">
<h1>Hello CodeSandbox</h1>
<h2>Start editing to see some magic happen!</h2>
<OpenModalButton>Modal A</OpenModalButton>
<OpenModalButton>Modal B</OpenModalButton>
</div>
</ModalProvider>
);
}
Demo

React useEffect Problems

So here is the problem which I can't seem to solve. I have an app component, inside of App I have rendered the Show Component. Show component has toggle functionality as well as a outside Click Logic. In the Show component I have a Button which removes an item based on his Id, problem is that When I click on the button Remove. It removes the item but it also closes the Show Component, I don't want that, I want when I press on button it removes the item but does not close the component. Thanks
App.js
const App =()=>{
const[isShowlOpen, setIsShowOpen]=React.useState(false)
const Show = useRef(null)
function openShow(){
setIsShowOpen(true)
}
function closeShowl(){
setIsShowOpen(false)
}
const handleShow =(e)=>{
if(show.current&& !showl.current.contains(e.target)){
closeShow()
}
}
useEffect(()=>{
document.addEventListener('click',handleShow)
return () =>{
document.removeEventListener('click', handleShow)
}
},[])
return (
<div>
<div ref={show}>
<img className='taskbar__iconsRight' onClick={() =>
setIsShowOpen(!isShowOpen)}
src="https://winaero.com/blog/wp-content/uploads/2017/07/Control-
-icon.png"/>
{isShowOpen ? <Show closeShow={closeShow} />: null}
</div>
)
}
Show Component
import React, { useContext } from 'react'
import './Show.css'
import { useGlobalContext } from '../../context'
import WindowsIcons from '../../WindowsIcons/WindowsIcons'
import { GrClose } from 'react-icons/gr'
const Show = ({closeShow}) => {
const {remove, icons }= useGlobalContext()
}
return (
<div className='control__Panel'>
<div className='close__cont'>
<GrClose className='close' onClick={closeShow} />
<h3>Show</h3>
</div>
<div className='control__cont'>
{icons.map((unin)=>{
const { name, img, id} = unin
return (
<li className='control' key ={id}>
<div className='img__text'>
<img className='control__Img' src={img} />
<h4 className='control__name'>{name}</h4>
</div>
<button className='unin__button' onClick={() => remove(id)} >remove</button>
</li> )
})}
</div>
</div>
)
}
export default Show

Try stop propagation function, it should stop the propagation of the click event
<button
className='unin__button'
onClick={(e) => {
e.stopPropagation();
remove(id);
}}
>remove</button>

You have a few typos in your example. Are they in your code? If they are, you're always reach the closeShow() case in your handler, since you're using the wrong ref.
const App =()=>{
const[isShowOpen, setIsShowOpen]=React.useState(false) <<-- Here 'isShowlOpen'
const show = useRef(null) <<-- here 'S'how
function openShow(){
setIsShowOpen(true)
}
function closeShow(){ <<-- Here 'closeShowl'
setIsShowOpen(false)
}
const handleShow =(e)=>{
if(show.current&& !show.current.contains(e.target)){ <<-- here 'showl'
closeShow()
}
}
useEffect(()=>{
document.addEventListener('click',handleShow)
return () =>{
document.removeEventListener('click', handleShow)
}
},[])
return (
<div>
<div ref={show}>
<img className='taskbar__iconsRight' onClick={() =>
setIsShowOpen(!isShowOpen)}
src="https://winaero.com/blog/wp-content/uploads/2017/07/Control-
-icon.png"/>
{isShowOpen ? <Show closeShow={closeShow} />: null}
</div>
)
}

React render list only when data source changes

Basically I have a modal with a state in the parent component and I have a component that renders a list. When I open the modal, I dont want the list to re render every time because there can be hundreds of items in the list its too expensive. I only want the list to render when the dataSource prop changes.
I also want to try to avoid using useMemo if possible. Im thinking maybe move the modal to a different container, im not sure.
If someone can please help it would be much appreciated. Here is the link to sandbox: https://codesandbox.io/s/rerender-reactmemo-rz6ss?file=/src/App.js

Since you said you want to avoid React.memo, I think the best approach would be to move the <Modal /> component to another "module"
export default function App() {
return (
<>
<Another list={list} />
<List dataSource={list} />
</>
);
}
And inside <Another /> component you would have you <Modal />:
import React, { useState } from "react";
import { Modal } from "antd";
const Another = ({ list }) => {
const [showModal, setShowModal] = useState(false);
return (
<div>
<Modal
visible={showModal}
onCancel={() => setShowModal(false)}
onOk={() => {
list.push({ name: "drink" });
setShowModal(false);
}}
/>
<button onClick={() => setShowModal(true)}>Show Modal</button>
</div>
)
}
export default Another
Now the list don't rerender when you open the Modal

You can use React.memo, for more information about it please check reactmemo
const List = React.memo(({ dataSource, loading }) => {
console.log("render list");
return (
<div>
{dataSource.map((i) => {
return <div>{i.name}</div>;
})}
</div>
);
});
sandbox here

how to hide menu slider on clicking a body using react js?

I wanted to hide menu slider onclicking a body in reactjs. how can i do that in function using react.js.
document.querySelector('.open-menu').onclick = function(){
html.classList.add('active');
};
document.querySelector('.close-menu').onclick = function(){
html.classList.remove('active');
};
html.onclick = function(event){
if (event.target === html){
html.classList.remove('active');
}
};
I want this same functionality in react js.

Check the code below.
import React, { useState } from 'react';
const SomeComponent = () => {
const [isMenuOpen, showMenu] = useState(false)
const toggleMenu = () => showMenu(!isMenuOpen)
return (
<>
{isMenuOpen && <MenuCompoment />}
<div onClick={toggleMenu}><App /></div>
</>
)
}

This is a stripped down version of code I've used before.
UseEffect on mounting of the Menu adds an event listener on the document for the click event.
When a click happens it uses closest to look up the parent tree of elements for an id (note the '#')
If it finds one, then the click happened on the menu otherwise it happened on any other element so close.
When the menu is disposed the return function of useEffect is called and removes the event handler.
import React, {useState, useEffect} from 'react';
const Page = () => {
const [toggle, setToggle] = useState(false);
return <div>
<button type="button" onClick={e => setToggle(!toggle)}>Toggle</button>
{ toggle && <Menu show={toggle} hide={() => setToggle(false)}/>}
</div>
}
const Menu = ({show, hide}) => {
useEffect(() => {
document.addEventListener("click", listen);
return () => {
document.removeEventListener("click", listen);
}
}, []);
const listen = e => {
if (!e.target.closest("#menu")) {
hide();
}
}
return <div className="menu" id="menu">
<span>I'm a menu</span>
</div>;
}

i think setting onclick event on the menuItems like this will Work
onClick={()=> setOpen(!open)}
export function SidebarMenu({open, setOpen}) {
return (
<div open={open}>
<Link to="#" title="Home" onClick={()=> setOpen(!open)} >Home</Link>
</div>
)
}

Probably too late for answer but since I saw it in active feed, I will try my best to answer it.
I can see that if your menu is open, you want to hide it if clicked anywhere else. I have used useRef to store the menu node and I compare it to the document whenever its open, if it is, I close the menu
Codesandbox link