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>
  )
}

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)

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
    })
}

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']);

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。