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

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

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}
        );
}

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);
componentDidMount(){
    //http://localhost:3000/service/anas
    //<Route path="/service/:serviceName" component={Service} />
    const {params} =this.props.match;
    this.setState({ 
        title: params.serviceName ,
        content: data.Content
    })
}

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

在React-Router-Dom V5中

function useQeury() {
 const [query, setQeury] = useState({});
 const search = useLocation().search.slice(1);

 useEffect(() => {
   setQeury(() => {
     const query = new URLSearchParams(search);
     const result = {};
     for (let [key, value] of query.entries()) {
       result[key] = value;
     }
     setQeury(result);
   }, [search]);
 }, [search, setQeury]);

 return { ...query };
}


// you can destruct query search like:
const {page , search} = useQuery()

// result
// {page : 1 , Search: "ABC"}