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

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

其他回答


最简单的解决方案!

在路由方面:

   <Route path="/app/someUrl/:id" exact component={binder} />

在react代码中:

componentDidMount() {
    var id = window.location.href.split('/')[window.location.href.split('/').length - 1];
    var queryString = "http://url/api/controller/" + id
    $.getJSON(queryString)
      .then(res => {
        this.setState({ data: res });
      });
  }

React Router v4不再有props.location.query对象(见github讨论)。因此,已接受的答案将不适用于较新的项目。

v4的解决方案是使用外部库查询字符串来解析props.location.search

const qs = require('query-string');
//or
import * as qs from 'query-string';

console.log(location.search);
//=> '?foo=bar'

const parsed = qs.parse(location.search);
console.log(parsed);
//=> {foo: 'bar'}

使用let {redirectParam} = useParams();如果你用的是功能组件

它是一个类组件

constructor (props) {  
        super(props);
        console.log(props);
        console.log(props.match.params.redirectParam)
}
async componentDidMount(){ 
        console.log(this.props.match.params.redirectParam)
}

React路由器v4

const urlParams = new URLSearchParams(this.props.location.search)
const key = urlParams.get('__firebase_request_key')

请注意,它目前还处于试验阶段。

查看浏览器兼容性:https://developer.mozilla.org/en-US/docs/Web/API/URLSearchParams/URLSearchParams#Browser_compatibility

在typescript中,参见下面的示例片段:

const getQueryParams = (s?: string): Map<string, string> => {
  if (!s || typeof s !== 'string' || s.length < 2) {
    return new Map();
  }

  const a: [string, string][] = s
    .substr(1) // remove `?`
    .split('&') // split by `&`
    .map(x => {
      const a = x.split('=');
      return [a[0], a[1]];
    }); // split by `=`

  return new Map(a);
};

在react中使用react-router-dom,你可以做

const {useLocation} from 'react-router-dom';
const s = useLocation().search;
const m = getQueryParams(s);

参见下面的例子

//下面是上面转换和缩小的ts函数 如果(const getQueryParams = t = > {! t | |“字符串”!=typeof t||t.length<2)return new Map;const r=t.substr(1).split("&")。地图(t = > {const r = t.split(" = ");返回[r[0],[1]]});返回新地图(r)}; //一个示例查询字符串 Const s = '?__arg1 = value1&arg2 = value2 ' getQueryParams(s) console.log (m.get (__arg1)) console.log (m.get(最长)) Console.log (m.t get('arg3')) //不存在,返回undefined