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

React路由器Dom V6 https://reactrouter.com/docs/en/v6/hooks/use-search-params

import * as React from "react";
import { useSearchParams } from "react-router-dom";

function App() {
  let [searchParams, setSearchParams] = useSearchParams();

  function handleSubmit(event) {
    event.preventDefault();
    // The serialize function here would be responsible for
    // creating an object of { key: value } pairs from the
    // fields in the form that make up the query.
    let params = serializeFormQuery(event.target);
    setSearchParams(params);
  }

  return (
    <div>
      <form onSubmit={handleSubmit}>{/* ... */}</form>
    </div>
  );
}

直到React路由器Dom V5

function useQueryParams() {
    const params = new URLSearchParams(
      window ? window.location.search : {}
    );

    return new Proxy(params, {
        get(target, prop) {
            return target.get(prop)
        },
    });
}

React钩子很棒

如果你的url看起来像/users?页面= 2数= 10字段=姓名、电子邮件、电话

// app.domain.com/users?page=2&count=10&fields=name,email,phone

const { page, fields, count, ...unknown } = useQueryParams();

console.log({ page, fields, count })
console.log({ unknown })

如果您的查询参数包含hyphone("-")或空格(" ") 然后你不能像{page, fields, count，…未知的}

你需要做传统的作业，比如

// app.domain.com/users?utm-source=stackOverFlow

const params = useQueryParams();

console.log(params['utm-source']);

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))

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" }

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}));
}

React路由器v6

来源:在React路由器中获取查询字符串(搜索参数)

使用新的useSearchParams钩子和.get()方法:

const Users = () => {
  const [searchParams] = useSearchParams();
  console.log(searchParams.get('sort')); // 'name'

  return <div>Users</div>;
};

使用这种方法，您可以读取一个或几个参数。

将参数作为一个对象:

如果你需要一次性获得所有的查询字符串参数，那么我们可以像这样使用Object.fromEntries:

const Users = () => {
  const [searchParams] = useSearchParams();
  console.log(Object.fromEntries([...searchParams])); // ▶ { sort: 'name', order: 'asecnding' }
  return <div>Users</div>;
};

阅读更多和现场演示:在React路由器中获取查询字符串(搜索参数)