如何在我的路由中定义路由。jsx文件捕获__firebase_request_key参数值从一个URL生成的Twitter的单点登录过程后，从他们的服务器重定向?
http://localhost:8000/#/signin?_k=v9ifuf&__firebase_request_key=blablabla
我尝试了以下路由配置，但:redirectParam没有捕获提到的参数:
<Router>
<Route path="/" component={Main}>
<Route path="signin" component={SignIn}>
<Route path=":redirectParam" component={TwitterSsoButton} />
</Route>
</Route>
</Router>
当前回答
http://localhost:8000/#/signin?id=12345
import React from "react";
import { useLocation } from "react-router-dom";
const MyComponent = () => {
const search = useLocation().search;
const id=new URLSearchParams(search).get("id");
console.log(id);//12345
}
其他回答
试试这个
http://localhost:4000/#/amoos?id=101
// ReactJS
import React from "react";
import { useLocation } from "react-router-dom";
const MyComponent = () => {
const search = useLocation().search;
const id = new URLSearchParams(search).get("id");
console.log(id); //101
}
// VanillaJS
const id = window.location.search.split("=")[1];
console.log(id); //101
我使用了一个名为query-string的外部包来解析url参数，如下所示。
import React, {Component} from 'react'
import { parse } from 'query-string';
resetPass() {
const {password} = this.state;
this.setState({fetching: true, error: undefined});
const query = parse(location.search);
return fetch(settings.urls.update_password, {
method: 'POST',
headers: {'Content-Type': 'application/json', 'Authorization': query.token},
mode: 'cors',
body: JSON.stringify({password})
})
.then(response=>response.json())
.then(json=>{
if (json.error)
throw Error(json.error.message || 'Unknown fetch error');
this.setState({fetching: false, error: undefined, changePassword: true});
})
.catch(error=>this.setState({fetching: false, error: error.message}));
}
如果你没有得到这个。道具…根据其他答案，您可能需要使用withthrouter (docs v4):
import React from 'react'
import PropTypes from 'prop-types'
import { withRouter } from 'react-router'
// A simple component that shows the pathname of the current location
class ShowTheLocation 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>You are now at {location.pathname}</div>
)
}
}
// Create a new component that is "connected" (to borrow redux terminology) to the router.
const TwitterSsoButton = withRouter(ShowTheLocation)
// This gets around shouldComponentUpdate
withRouter(connect(...)(MyComponent))
// This does not
connect(...)(withRouter(MyComponent))
容易解构分配URLSearchParams
测试尝试如下:
1 扫描:https://www.google.com/?param1=apple¶m2=banana
2 右键单击>页，单击Inspect > goto Console选项卡 然后粘贴下面的代码:
const { param1, param2 } = Object.fromEntries(new URLSearchParams(location.search));
console.log("YES!!!", param1, param2 );
输出:
YES!!! apple banana
你可以扩展params，如param1, param2，想扩展多少就扩展多少。
如果你的路由器是这样的
<Route exact path="/category/:id" component={ProductList}/>
你会得到这样的id
this.props.match.params.id
