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
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'}
从v4开始，React路由器不再直接在其location对象中提供查询参数。原因是
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".
包含了这个之后，只解析location会更有意义。在需要查询对象的视图组件中搜索。
你可以通过覆盖react-router中的withRouter来实现这一点
customWithRouter.js
import { compose, withPropsOnChange } from 'recompose';
import { withRouter } from 'react-router';
import queryString from 'query-string';
const propsWithQuery = withPropsOnChange(
['location', 'match'],
({ location, match }) => {
return {
location: {
...location,
query: queryString.parse(location.search)
},
match
};
}
);
export default compose(withRouter, propsWithQuery)
在没有第三方库或复杂的解决方案的情况下，在一行中完成这一切。以下是如何
let myVariable = new URLSearchParams(history.location.search).get('business');
你唯一需要改变的是你自己的参数名称的单词“business”。
业务= url.com例子吗?你好
myVariable的结果将是hello
或者像这样?
Let win = { “位置”:{ “路径”:“http://localhost: 8000 / # / signin吗?_k = v9ifuf&__firebase_request_key =之类的 ｝ ｝ If (win.location.path.match('__firebase_request_key').length) { 让key= win.location.path.split('__firebase_request_key=')[1] console.log(关键) ｝
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);
