This is my main section class where all routes to the links is kept
export default class Section extends React.Component { render() { return ( <div style={{backgroundColor:"white"}}> <Routes> <Route path="/" element={<Home/>} /> <Route path="/discover" element={<Discover/>} /> <Route path="/shop/makeup" element={<Makeup/>} /> <Route path="/home/:id" element={<DetailsPage/>} /> </Routes> </div> )}}
This is my card page which is getting data from the context.
import React from 'react'; import {DataContext} from './CardData.js'; import {Link} from 'react-router-dom' import '../App.css'; export default class HomeCard extends React.Component { static contextType = DataContext; render(){ const {products} = this.context; return ( <div> <div className="card"> { products.map((val,index)=>{ return( <div className="card" key={val.id}> <Card style={{ width: '14rem' }} className="cardhead"> <Card.Img variant="top" src={val.imgsrc} className="cardimg"/> <Card.Body> <Card.Text>{val.mname} </Card.Text> <Card
From here i had passed the val.id to the url of the page using LINK
<Link to={`/home/${val.id}`}> <Button className="overlay" variant="primary"> {/* <a style={{color:"white"}} href={props.link} className="card-link">View</a> */} View</Button> </Link> <Card.Text><strong>{val.price}</strong> </Card.Text> <Card.Text><strong>{val.id}</strong> </Card.Text> </Card.Footer> </Card.Body> </Card> </div>); </div> )}}
I want to access the the link url into the details page of my product which is as follows :
export default class DetailsPage extends React.Component { static contextType = DataContext; state = { product: [] } getProduct = () =>{ if(this.props.match.params.id){ const res = this.context.products; const data = res.filter(item =>{ return item.id === this.props.match.params.id }) this.setState({product: data}) }}; componentDidMount(){ this.getProduct(); } render() { const {product} = this.state; const {addCart} = this.context; return ( <> {product.map(item =>( <div className="details" key={item.id}> <img src={item.imgsrc} alt=""/> <div className="box"> <div className="row"> <h2>{item.mname}</h2> <span>${item.price}</span> </div> <Link to="/cart" className="cart" onClick={() => addCart(item.id)}> Add to cart </Link> </div> </div> ))} </> ) }}
Unfortunately it is giving an error saying TypeError: Cannot read property ‘params’ of undefined
Answers:
Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.
Method 1
Issue(s)
- react-router-dom v6
Route
components rendered via theelement
prop don’t receive route props. - Route children components must use react hooks to access the route context, i.e.
useParams
,useLocation
,useNavigate
, etc… and therefore must be function components. - There no longer exists a
withRouter
Higher Order Component.
Solution
DetailsPage
is a class-based component so I see a couple options for getting access to the route’s match params.
- Convert
DetailsPage
to be a function component and use theuseParams
react hook. - Write your own
withRouter
HOC to access the route context and pass as props any of the react-hook accessible values.
I won’t cover converting a class-based component to a function component, but can provide a simple HOC solution to pass to DetailsPage
(or any other class-based component) the match params.
const withRouter = WrappedComponent => props => { const params = useParams(); // etc... other react-router-dom v6 hooks return ( <WrappedComponent {...props} params={params} // etc... /> ); };
You can now wrap & export your class-based components in the very familiar way they were from v4/v5.
Method 2
Pass props via Link component in React Router v6 by a separate prop called state
like <Link to="/" state={{ foo: bar }}>
.
Example:
<Link to={`/login`} state={{ from: "the-page-id" }} > Login to see this page </Link>
And retrieve it using useLocation()
hook:
import { useLocation } from "react-router-dom"; ... const location = useLocation(); const { from } = location.state; console.log(from); // output: "the-page-id"
This should be used with function components only not class components.
Method 3
Using react-router-dom v6
Parent component (App)
<Route path="/something/:id" element={<Something />} />
Child component (Something)
import { useParams } from "react-router-dom"; const Something = (props) => { let { id } = useParams(); useEffect(() => { console.log(`/something/${id}`); }, []); // ..... }
Method 4
I had this same problem and scoured the internet for an easy solution. I didn’t want to rewrite all my class components as functional components but I thought I might have to. However, I used a withRouter function with useParams() and it works really well and is easy to follow.
To solve your problem so you don’t have to rewrite your DetailsPage class as a function, follow these steps:
-
Add the following line of code at the beginning of your class:
import { useParams } from 'react-router-dom';
-
Then add this function above your class (copy it exactly):
export function withRouter(Children){ return(props)=>{ const match = {params: useParams()}; return <Children {...props} match = {match}/> } }
-
Next, change your class definition to this:
class DetailsPage extends React.Component {
-
Add the following line of code at the end of your class:
export default withRouter(DetailsPage);
Method 5
const navigate = useNavigate();
function _navigateToPage (pageNumber) { const page = pageNumber const title = "Hello World"; navigate("/fragment", { state:{ page, title }, }); }
Retrieve parameters in other page with useLocation
const params = useLocation();
console.log(params);
All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0