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

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"}

其他回答

this.props.params。Your_param_name将工作。

这是从查询字符串中获取参数的方法。 请执行console.log(this.props);探索所有的可能性。

有了这一行代码,你可以在React Hook和React Class Component的任何地方使用它。

https://www.hunterisgod.com/?city=Leipzig

let city = (new URLSearchParams(window.location.search)).get("city")

容易解构分配URLSearchParams

测试尝试如下:

1 扫描:https://www.google.com/?param1=apple&param2=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。

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
    })
}

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"}