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

React路由器v3

使用React Router v3,你可以从this.props.location.search (?qs1=naisarg&qs2=parmar)获取查询字符串。例如,使用let params = queryString.parse(this.props.location.search),将给出{qs1: 'naisarg', qs2: 'parmar'}

React路由器v4

在React Router v4中,this.props.location.query不再存在。您需要使用this.props.location.search,并自己或使用现有的包(如query-string)解析查询参数。

例子

下面是一个使用React Router v4和query-string库的最小示例。

import { withRouter } from 'react-router-dom';
import queryString from 'query-string';
    
class ActivateAccount extends Component{
    someFunction(){
        let params = queryString.parse(this.props.location.search)
        ...
    }
    ...
}
export default withRouter(ActivateAccount);

理性的

React Router团队移除query属性的理由是:

There are a number of popular packages that do query string parsing/stringifying slightly differently, and each of these differences might be the "correct" way for some users and "incorrect" for others. If React Router picked the "right" one, it would only be right for some people. Then, it would need to add a way for other users to substitute in their preferred query parsing package. There is no internal use of the search string by React Router that requires it to parse the key-value pairs, so it doesn't have a need to pick which one of these should be "right". [...] The approach being taken for 4.0 is to strip out all the "batteries included" kind of features and get back to just basic routing. If you need query string parsing or async loading or Redux integration or something else very specific, then you can add that in with a library specifically for your use case. Less cruft is packed in that you don't need and you can customize things to your specific preferences and needs.

你可以在GitHub上找到完整的讨论。

其他回答

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>;
}

在需要访问可以使用的参数的组件中

this.props.location.state.from.search

这将显示整个查询字符串(在?标志)

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
}

也许有人可以帮助解释为什么,但如果你试图从App.js页面上的新安装的Create React App中点击props来查找位置,你会得到:

无法读取未定义的属性“搜索”

即使我有App.js作为主路径:

<Route exact path='/' render={props => (

只在App.js上,使用window。地点对我来说很合适:

import queryString from 'query-string';
...
const queryStringParams = queryString.parse(window.location.search);

你也可以使用react-location-query包,例如:

  const [name, setName] = useLocationField("name", {
    type: "string",
    initial: "Rostyslav"
  });

  return (
    <div className="App">
      <h1>Hello {name}</h1>
      <div>
        <label>Change name: </label>
        <input value={name} onChange={e => setName(e.target.value)} />
      </div>
    </div>
  );

名称-获取价值 setName =设置值

这个包有很多选项,在Github上的文档中阅读更多