我有几个按钮作为路径。每次改变路线时,我都想确保激活的按钮发生了变化。
有没有办法在react路由器v4中监听路由变化?
我有几个按钮作为路径。每次改变路线时,我都想确保激活的按钮发生了变化。
有没有办法在react路由器v4中监听路由变化?
当前回答
v5.1引入了有用的钩子useLocation
https://reacttraining.com/blog/react-router-v5-1/#uselocation
import { Switch, useLocation } from 'react-router-dom'
function usePageViews() {
let location = useLocation()
useEffect(
() => {
ga.send(['pageview', location.pathname])
},
[location]
)
}
function App() {
usePageViews()
return <Switch>{/* your routes here */}</Switch>
}
其他回答
import React, { useEffect } from 'react';
import { useLocation } from 'react-router';
function MyApp() {
const location = useLocation();
useEffect(() => {
console.log('route has been changed');
...your code
},[location.pathname]);
}
用钩子
使用useEffect钩子,可以在不添加侦听器的情况下检测路由更改。
import React, { useEffect } from 'react';
import { Switch, Route, withRouter } from 'react-router-dom';
import Main from './Main';
import Blog from './Blog';
const App = ({history}) => {
useEffect( () => {
// When route changes, history.location.pathname changes as well
// And the code will execute after this line
}, [history.location.pathname]);
return (<Switch>
<Route exact path = '/' component = {Main}/>
<Route exact path = '/blog' component = {Blog}/>
</Switch>);
}
export default withRouter(App);
v5.1引入了有用的钩子useLocation
https://reacttraining.com/blog/react-router-v5-1/#uselocation
import { Switch, useLocation } from 'react-router-dom'
function usePageViews() {
let location = useLocation()
useEffect(
() => {
ga.send(['pageview', location.pathname])
},
[location]
)
}
function App() {
usePageViews()
return <Switch>{/* your routes here */}</Switch>
}
对于功能组件,请尝试使用props.location中的useEffect。
import React, {useEffect} from 'react';
const SampleComponent = (props) => {
useEffect(() => {
console.log(props.location);
}, [props.location]);
}
export default SampleComponent;
您应该使用history v4 lib。
这里的例子
history.listen((location, action) => {
console.log(`The current URL is ${location.pathname}${location.search}${location.hash}`)
console.log(`The last navigation action was ${action}`)
})