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
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
})
}
其他回答
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")
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);
最简单的解决方案!
在路由方面:
<Route path="/app/someUrl/:id" exact component={binder} />
在react代码中:
componentDidMount() {
var id = window.location.href.split('/')[window.location.href.split('/').length - 1];
var queryString = "http://url/api/controller/" + id
$.getJSON(queryString)
.then(res => {
this.setState({ data: res });
});
}
在React Router v4中，只有withRoute才是正确的方式
您可以通过withRouter高阶组件访问历史对象的属性和最近的匹配。withRouter将在包装组件呈现时将更新的匹配、位置和历史道具传递给它。
import React from 'react'
import PropTypes from 'prop-types'
import { withRouter } from 'react-router'
// A simple component that shows the pathname of the current location
class ShowTheLocation extends React.Component {
static propTypes = {
match: PropTypes.object.isRequired,
location: PropTypes.object.isRequired,
history: PropTypes.object.isRequired
}
render() {
const { match, location, history } = this.props
return (
<div>You are now at {location.pathname}</div>
)
}
}
// Create a new component that is "connected" (to borrow redux
// terminology) to the router.
const ShowTheLocationWithRouter = withRouter(ShowTheLocation)
https://reacttraining.com/react-router/web/api/withRouter
不是反应的方式，但我相信这个单行函数可以帮助你:)
const getQueryParams = (query = null) => [...(new URLSearchParams(query||window.location.search||"")).entries()].reduce((a,[k,v])=>(a[k]=v,a),{});
或:
const getQueryParams = (query = null) => (query||window.location.search.replace('?','')).split('&').map(e=>e.split('=').map(decodeURIComponent)).reduce((r,[k,v])=>(r[k]=v,r),{});
或完整版:
const getQueryParams = (query = null) => {
return (
(query || window.location.search.replace("?", ""))
// get array of KeyValue pairs
.split("&")
// Decode values
.map((pair) => {
let [key, val] = pair.split("=");
return [key, decodeURIComponent(val || "")];
})
// array to object
.reduce((result, [key, val]) => {
result[key] = val;
return result;
}, {})
);
};
例子: URL:…?= = 1 b =建发集团有限公司的测试 代码:
getQueryParams()
//=> {a: "1", b: "c", d: "test"}
getQueryParams('type=user&name=Jack&age=22')
//=> {type: "user", name: "Jack", age: "22" }
