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路由器v3

使用React Router v3,你可以从this.props.location.search (?qs1=naisarg&qs2=parmar)获取查询字符串。例如,使用let params = queryString.parse(this.props.location.search),将给出{qs1: 'naisarg', qs2: 'parmar'}

React路由器v4

在React Router v4中,this.props.location.query不再存在。您需要使用this.props.location.search,并自己或使用现有的包(如query-string)解析查询参数。

例子

下面是一个使用React Router v4和query-string库的最小示例。

import { withRouter } from 'react-router-dom';
import queryString from 'query-string';
    
class ActivateAccount extends Component{
    someFunction(){
        let params = queryString.parse(this.props.location.search)
        ...
    }
    ...
}
export default withRouter(ActivateAccount);

理性的

React Router团队移除query属性的理由是:

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". [...] The approach being taken for 4.0 is to strip out all the "batteries included" kind of features and get back to just basic routing. If you need query string parsing or async loading or Redux integration or something else very specific, then you can add that in with a library specifically for your use case. Less cruft is packed in that you don't need and you can customize things to your specific preferences and needs.

你可以在GitHub上找到完整的讨论。

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")

你可以使用下面的react钩子:

如果url改变,钩子状态会更新 SSR: typeof window === "undefined",只是检查窗口导致错误(尝试一下) 代理对象隐藏实现,因此返回undefined而不是null

这是获取搜索参数为对象的函数:

const getSearchParams = <T extends object>(): Partial<T> => {
    // server side rendering
    if (typeof window === "undefined") {
        return {}
    }

    const params = new URLSearchParams(window.location.search) 

    return new Proxy(params, {
        get(target, prop, receiver) {
            return target.get(prop as string) || undefined
        },
    }) as T
}

然后像这样把它用作钩子:

const useSearchParams = <T extends object = any>(): Partial<T> => {
    const [searchParams, setSearchParams] = useState(getSearchParams())

    useEffect(() => {
        setSearchParams(getSearchParams())
    }, [typeof window === "undefined" ? "once" : window.location.search])

    return searchParams
}

如果你的url是这样的:

/app?page=2&count=10

你可以这样读:

const { page, count } = useQueryParams();

console.log(page, count)

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

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

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。