V6
You can use useNavigate
, useLocation
and useMatch
in your component to get matchPath
, navigate
and location
.
const Child = () => {
const location = useLocation();
const navigate = useNavigate();
const match = useMatch("write-the-url-you-want-to-match-here");
return (
<div>{location.pathname}</div>
)
}
export default Child
V5.1 & Hooks (Requires React >= 16.8)
You can use useHistory
, useLocation
and useRouteMatch
in your component to get match
, history
and location
.
const Child = () => {
const location = useLocation();
const history = useHistory();
const match = useRouteMatch("write-the-url-you-want-to-match-here");
return (
<div>{location.pathname}</div>
)
}
export default Child
V4 & V5
You can use withRouter
HOC in order to inject match
, history
and location
in your component props.
class Child extends React.Component {
static propTypes = {
match: PropTypes.object.isRequired,
location: PropTypes.object.isRequired,
history: PropTypes.object.isRequired
}
render() {
const { match, location, history } = this.props
return (
<div>{location.pathname}</div>
)
}
}
export default withRouter(Child)
V3
You can use withRouter
HOC in order to inject router
, params
, location
, routes
in your component props.
class Child extends React.Component {
render() {
const { router, params, location, routes } = this.props
return (
<div>{location.pathname}</div>
)
}
}
export default withRouter(Child)
Original answer
If you don’t want to use the props, you can use the context as described in React Router documentation
First, you have to set up your childContextTypes
and getChildContext
class App extends React.Component{
getChildContext() {
return {
location: this.props.location
}
}
render() {
return <Child/>;
}
}
App.childContextTypes = {
location: React.PropTypes.object
}
Then, you will be able to access to the location object in your child components using the context like this
class Child extends React.Component{
render() {
return (
<div>{this.context.location.pathname}</div>
)
}
}
Child.contextTypes = {
location: React.PropTypes.object
}