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 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'}
其他回答
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
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'}
您可以使用这段代码来获取作为对象的参数。如果url中没有查询参数，该对象将为空
let url = window.location.toString(); Let params = url?.split("?")[1]?.split("&"); 让obj = {}; params?.forEach((el) => { Let [k, v] = el?.split("="); obj[k] = v.replaceAll("%20"， " " "); })； console.log (obj);
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 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'}
