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

不是反应的方式,但我相信这个单行函数可以帮助你:)

const getQueryParams = (query = null) => [...(new URLSearchParams(query||window.location.search||"")).entries()].reduce((a,[k,v])=>(a[k]=v,a),{});

或:

const getQueryParams = (query = null) => (query||window.location.search.replace('?','')).split('&').map(e=>e.split('=').map(decodeURIComponent)).reduce((r,[k,v])=>(r[k]=v,r),{});

或完整版:

const getQueryParams = (query = null) => {
  return (
    (query || window.location.search.replace("?", ""))

      // get array of KeyValue pairs
      .split("&") 

      // Decode values
      .map((pair) => {
        let [key, val] = pair.split("=");

        return [key, decodeURIComponent(val || "")];
      })

      // array to object
      .reduce((result, [key, val]) => {
        result[key] = val;
        return result;
      }, {})
  );
};

例子: URL:…?= = 1 b =建发集团有限公司的测试 代码:

getQueryParams()
//=> {a: "1", b: "c", d: "test"}

getQueryParams('type=user&name=Jack&age=22')
//=> {type: "user", name: "Jack", age: "22" }

其他回答

export class ClassName extends Component{
      constructor(props){
        super(props);
        this.state = {
          id:parseInt(props.match.params.id,10)
        }
    }
     render(){
        return(
          //Code
          {this.state.id}
        );
}

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

在React Router v4中,只有withRoute才是正确的方式

您可以通过withRouter高阶组件访问历史对象的属性和最近的匹配。withRouter将在包装组件呈现时将更新的匹配、位置和历史道具传递给它。

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 ShowTheLocationWithRouter = withRouter(ShowTheLocation)

https://reacttraining.com/react-router/web/api/withRouter

React路由器5.1+

5.1引入了各种钩子,如useLocation和useParams,可以在这里使用。

例子:

<Route path="/test/:slug" component={Dashboard} />

如果我们去参观

http://localhost:3000/test/signin?_k=v9ifuf&__firebase_request_key=blablabla

你可以把它找回来

import { useLocation } from 'react-router';
import queryString from 'query-string';

const Dashboard: React.FC = React.memo((props) => {
    const location = useLocation();

    console.log(queryString.parse(location.search));

    // {__firebase_request_key: "blablabla", _k: "v9ifuf"}

    ...

    return <p>Example</p>;
}

当你使用react route dom时,将用for match清空对象,但如果你执行以下代码,则它将用于es6组件,以及它直接用于函数组件

import { Switch, Route, Link } from "react-router-dom";

<Route path="/profile" exact component={SelectProfile} />
<Route
  path="/profile/:profileId"
  render={props => {
    return <Profile {...props} loading={this.state.loading} />;
  }}
/>
</Switch>
</div>

通过这种方式,您可以获得道具并匹配参数和配置文件id

在对es6组件进行了大量研究后,这对我来说很有效。