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>
容易解构分配URLSearchParams
测试尝试如下:
1
扫描:https://www.google.com/?param1=apple¶m2=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。
你可以创建一个简单的钩子来从当前位置提取搜索参数:
import React from 'react';
import { useLocation } from 'react-router-dom';
export function useSearchParams<ParamNames extends string[]>(...parameterNames: ParamNames): Record<ParamNames[number], string | null> {
const { search } = useLocation();
return React.useMemo(() => { // recalculate only when 'search' or arguments changed
const searchParams = new URLSearchParams(search);
return parameterNames.reduce((accumulator, parameterName: ParamNames[number]) => {
accumulator[ parameterName ] = searchParams.get(parameterName);
return accumulator;
}, {} as Record<ParamNames[number], string | null>);
}, [ search, parameterNames.join(',') ]); // join for sake of reducing array of strings to simple, comparable string
}
然后你可以像这样在你的功能组件中使用它:
// current url: http://localhost:8000/#/signin?_k=v9ifuf&__firebase_request_key=blablabla
const { __firebase_request_key } = useSearchParams('__firebase_request_key');
// current url: http://localhost:3000/home?b=value
const searchParams = useSearchParameters('a', 'b'); // {a: null, b: 'value'}