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

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

如果你没有得到这个。道具…根据其他答案，您可能需要使用withthrouter (docs v4):

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 TwitterSsoButton = withRouter(ShowTheLocation)  

// This gets around shouldComponentUpdate
withRouter(connect(...)(MyComponent))

// This does not
connect(...)(withRouter(MyComponent))

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)

容易解构分配URLSearchParams

测试尝试如下:

1 扫描:https://www.google.com/?param1=apple&param2=banana

2 右键单击>页，单击Inspect > goto Console选项卡 然后粘贴下面的代码:

const { param1, param2 } = Object.fromEntries(new URLSearchParams(location.search));
console.log("YES!!!", param1, param2 );

输出:

YES!!! apple banana

你可以扩展params，如param1, param2，想扩展多少就扩展多少。

我使用了一个名为query-string的外部包来解析url参数，如下所示。

import React, {Component} from 'react'
import { parse } from 'query-string';

resetPass() {
    const {password} = this.state;
    this.setState({fetching: true, error: undefined});
    const query = parse(location.search);
    return fetch(settings.urls.update_password, {
        method: 'POST',
        headers: {'Content-Type': 'application/json', 'Authorization': query.token},
        mode: 'cors',
        body: JSON.stringify({password})
    })
        .then(response=>response.json())
        .then(json=>{
            if (json.error)
                throw Error(json.error.message || 'Unknown fetch error');
            this.setState({fetching: false, error: undefined, changePassword: true});
        })
        .catch(error=>this.setState({fetching: false, error: error.message}));
}