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

据我所知,有三种方法可以做到。

1.使用正则表达式获取查询字符串。

2.您可以使用浏览器api。 图片当前的url是这样的:

http://www.google.com.au?token=123

我们只想得到123;

第一个

 const query = new URLSearchParams(this.props.location.search);

Then

const token = query.get('token')
console.log(token)//123

使用第三个名为“query-string”的库。 首先安装它 NPM I查询字符串 然后导入到当前的javascript文件中: 导入query-string

下一步是在当前url中获取'token',请执行以下操作:

const value=queryString.parse(this.props.location.search);
const token=value.token;
console.log('token',token)//123

2019年2月25日更新

4. 如果当前url如下所示:

http://www.google.com.au?app=home&act=article&aid=160990

我们定义一个函数来获取参数:

function getQueryVariable(variable)
{
        var query = window.location.search.substring(1);
        console.log(query)//"app=article&act=news_content&aid=160990"
        var vars = query.split("&");
        console.log(vars) //[ 'app=article', 'act=news_content', 'aid=160990' ]
        for (var i=0;i<vars.length;i++) {
                    var pair = vars[i].split("=");
                    console.log(pair)//[ 'app', 'article' ][ 'act', 'news_content' ][ 'aid', '160990' ] 
        if(pair[0] == variable){return pair[1];}
         }
         return(false);
}

我们可以通过以下方式获得“援助”:

getQueryVariable('aid') //160990

其他回答

也许有点晚了,但是这个react钩子可以帮助你在URL查询中获取/设置值:https://github.com/rudyhuynh/use-url-search-params(由我编写)。

不管有没有反应路由器,它都可以工作。 下面是您案例中的代码示例:

import React from "react";
import { useUrlSearchParams } from "use-url-search-params";

const MyComponent = () => {
  const [params, setParams] = useUrlSearchParams()
  return (
    <div>
      __firebase_request_key: {params.__firebase_request_key}
    </div>
  )
}

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'}

在typescript中,参见下面的示例片段:

const getQueryParams = (s?: string): Map<string, string> => {
  if (!s || typeof s !== 'string' || s.length < 2) {
    return new Map();
  }

  const a: [string, string][] = s
    .substr(1) // remove `?`
    .split('&') // split by `&`
    .map(x => {
      const a = x.split('=');
      return [a[0], a[1]];
    }); // split by `=`

  return new Map(a);
};

在react中使用react-router-dom,你可以做

const {useLocation} from 'react-router-dom';
const s = useLocation().search;
const m = getQueryParams(s);

参见下面的例子

//下面是上面转换和缩小的ts函数 如果(const getQueryParams = t = > {! t | |“字符串”!=typeof t||t.length<2)return new Map;const r=t.substr(1).split("&")。地图(t = > {const r = t.split(" = ");返回[r[0],[1]]});返回新地图(r)}; //一个示例查询字符串 Const s = '?__arg1 = value1&arg2 = value2 ' getQueryParams(s) console.log (m.get (__arg1)) console.log (m.get(最长)) Console.log (m.t get('arg3')) //不存在,返回undefined

你也可以使用react-location-query包,例如:

  const [name, setName] = useLocationField("name", {
    type: "string",
    initial: "Rostyslav"
  });

  return (
    <div className="App">
      <h1>Hello {name}</h1>
      <div>
        <label>Change name: </label>
        <input value={name} onChange={e => setName(e.target.value)} />
      </div>
    </div>
  );

名称-获取价值 setName =设置值

这个包有很多选项,在Github上的文档中阅读更多

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)