我似乎找不到如何更新查询参数与反应路由器不使用<Link/>。hashHistory.push(url)似乎没有注册查询参数,而且似乎不能将查询对象或任何东西作为第二个参数传递。

如何将url从/shop/Clothes/dresses更改为/shop/Clothes/dresses?color=blue在反应路由器没有使用<链接>?

onChange函数真的是侦听查询更改的唯一方法吗?为什么不自动检测和响应查询更改-以参数更改的方式?


当前回答

我目前在一个正在运行的项目中使用react-router v5,不容易迁移到v6。 我写了一个钩子,允许读取和修改一个URL参数,同时保持其他URL参数不变。 数组被视为逗号分隔值的列表: magnifying_glass ?产品=管,猎鹿帽

import { useCallback } from 'react';
import { useHistory } from 'react-router';

const getDecodedUrlParam = (name: string, locationSearch: string, _default?: any) => {
  const params = deserialize(locationSearch);
  const param = params[name];

  if (_default && Array.isArray(_default)) {
    return param
      ? param.split(',').map((v: string) => decodeURIComponent(v))
      : _default;
  }

  return param ? decodeURIComponent(param) : _default;
};

const deserialize = (locationSearch: string): any => {
  if (locationSearch.startsWith('?')) {
    locationSearch = locationSearch.substring(1);
  }
  const parts = locationSearch.split('&');
  return Object.fromEntries(parts.map((part) => part.split('=')));
};

const serialize = (params: any) =>
  Object.entries(params)
    .map(([key, value]) => `${key}=${value}`)
    .join('&');

export const useURLSearchParam = (name: string, _default?: any) => {
  const history = useHistory();
  const value: any = getDecodedUrlParam(name, location.search, _default);
  const _update = useCallback(
    (value: any) => {
      const params = deserialize(location.search);
      if (Array.isArray(value)) {
        params[name] = value.map((v) => encodeURIComponent(v)).join(',');
      } else {
        params[name] = encodeURIComponent(value);
      }
      history.replace({ pathname: location.pathname, search: serialize(params) });
    },
    [history, name]
  );
  const _delete = useCallback(() => {
    const params = deserialize(location.search);
    delete params[name];
    history.replace({ pathname: location.pathname, search: serialize(params) });
  }, [history, name]);
  return [value, _update, _delete];
};

其他回答

在hashHistory的push方法中,可以指定查询参数。例如,

history.push({
  pathname: '/dresses',
  search: '?color=blue'
})

or

history.push('/dresses?color=blue')

您可以查看这个存储库以获得有关使用历史记录的其他示例

当你需要一个模块来轻松地解析你的查询字符串时,推荐使用query-string模块。

http://localhost:3000?token=xxx-xxx-xxx

componentWillMount() {
    var query = queryString.parse(this.props.location.search);
    if (query.token) {
        window.localStorage.setItem("jwt", query.token);
        store.dispatch(push("/"));
    }
}

在这里,我从Node.js服务器重定向回我的客户端后,成功的Google-Passport身份验证,这是重定向回令牌作为查询参数。

我用query-string模块解析它,保存它并更新url中的查询参数,从react-router-redux推送。

就像@Craques解释的那样,我们可以使用替换功能,而不是每次更改都推送一个新路由。然而,在react-router的第6版中,useHistory()被useNavigate()取代,它只返回一个函数。你可以将选项传递给函数,以达到与旧的location.replace()相同的效果:

import { useLocation, useNavigate } from 'react-router-dom';
const to = { pathname: location.pathname, search: newParams.toString() };
navigate(to, { replace: true });

我目前在一个正在运行的项目中使用react-router v5,不容易迁移到v6。 我写了一个钩子,允许读取和修改一个URL参数,同时保持其他URL参数不变。 数组被视为逗号分隔值的列表: magnifying_glass ?产品=管,猎鹿帽

import { useCallback } from 'react';
import { useHistory } from 'react-router';

const getDecodedUrlParam = (name: string, locationSearch: string, _default?: any) => {
  const params = deserialize(locationSearch);
  const param = params[name];

  if (_default && Array.isArray(_default)) {
    return param
      ? param.split(',').map((v: string) => decodeURIComponent(v))
      : _default;
  }

  return param ? decodeURIComponent(param) : _default;
};

const deserialize = (locationSearch: string): any => {
  if (locationSearch.startsWith('?')) {
    locationSearch = locationSearch.substring(1);
  }
  const parts = locationSearch.split('&');
  return Object.fromEntries(parts.map((part) => part.split('=')));
};

const serialize = (params: any) =>
  Object.entries(params)
    .map(([key, value]) => `${key}=${value}`)
    .join('&');

export const useURLSearchParam = (name: string, _default?: any) => {
  const history = useHistory();
  const value: any = getDecodedUrlParam(name, location.search, _default);
  const _update = useCallback(
    (value: any) => {
      const params = deserialize(location.search);
      if (Array.isArray(value)) {
        params[name] = value.map((v) => encodeURIComponent(v)).join(',');
      } else {
        params[name] = encodeURIComponent(value);
      }
      history.replace({ pathname: location.pathname, search: serialize(params) });
    },
    [history, name]
  );
  const _delete = useCallback(() => {
    const params = deserialize(location.search);
    delete params[name];
    history.replace({ pathname: location.pathname, search: serialize(params) });
  }, [history, name]);
  return [value, _update, _delete];
};

您可以使用replace功能,而不是在每次更改时推送一个新路由

import React from 'react';
import { useHistory, useLocation } from 'react-router';

const MyComponent = ()=>{
   const history = useHistory();
   const location = useLocation();

   const onChange=(event)=>{
     const {name, value} = event?.target;
     const params = new URLSearchParams({[name]: value });
     history.replace({ pathname: location.pathname, search: params.toString() });       
   }

   return <input name="search" onChange={onChange} />
}

这保留了历史,而不是在每一个变化上都推一条新的道路

更新- 2022年2月(V6)

正如Matrix Spielt指出的那样,useHistory被usenavate取代来进行更改。还有一个方便的方法叫做useSearchParams,我只需要阅读文档,没有运行这个,但这应该可以工作

import React from 'react';
import { useSearchParams } from 'react-router-dom';
// import from react-router should also work but following docs
// import { useSearchParams } from 'react-router';

const MyComponent = ()=>{
   const [searchParams, setSearchParams] = useSearchParams();

   const onChange=(event)=>{
     const {name, value} = event?.target;
     setSearchParams({[name]: value})       
   }

   return <input name="search" onChange={onChange} />
}