从 url 中删除查询参数

Question

我有以下网址：

http://my.site/?code=74e30ef2-109c-4b75-b8d6-89bdce1aa860

我想重定向到：

http://my.site#/homepage

我这样做：

import { push } from 'react-router-redux'

...

dispatch(push('/homepage'))

但反应带我到：

http://my.site/?code=74e30ef2-109c-4b75-b8d6-89bdce1aa860#/homepage

如何在不重新加载应用程序的情况下告诉 React 将查询参数删除到浏览器的地址栏中？

Answer 1

使用正则表达式也许这个解决方案对你来说更容易：

  var oldURL = 'http://my.site/?code=74e30ef2-109c-4b75-b8d6-89bdce1aa860'
  var newURL = /(http(s?).+)(\?.+)/.exec(oldDomain)[1] + 'homepage'

您可以在项目中使用 newURL。

Answer 2

试试这个（硬编码方式）

import { push } from 'react-router-redux'

...
const url = `${window.location.origin}/homepage`;
dispatch(push(url));

或者这个

history.location.pathname =`${window.location.origin}/homepage`;

这些方法根本不是一个好的做法，而是那些工作（肯定是第二个）

阅读更多

Answer 3

根据以下链接，我 git 克隆了它并在主目录和示例基本目录中运行 npm install，然后运行 ​​npm start 以获得一个工作示例。

https://github.com/reactjs/react-router-redux

https://github.com/reactjs/react-router-redux/tree/master/examples/basic

https://github.com/reactjs/react-router-redux#pushlocation-replacelocation-gonumber-goback-goforward

删除查询参数（靠近底部）并验证它们已被删除的代码：

import { createStore, combineReducers, applyMiddleware } from 'redux'
import { browserHistory } from 'react-router'
import { syncHistoryWithStore, routerReducer, routerMiddleware, push } from 'react-router-redux'
import * as reducers from './reducers'

const reducer = combineReducers({
  ...reducers,
  routing: routerReducer
})

const middleware = routerMiddleware(browserHistory)

const store = createStore(
  reducer,
  applyMiddleware(middleware)
)
const history = syncHistoryWithStore(browserHistory, store)

// Dispatch from anywhere like normal.
store.dispatch(push("/?test=ok"));
// store2.dispatch(push(history.createHref("/")));
var count = 0;

history.listen(location => {
    console.log(location);
    if (count > 1) { return; }
    count++;
    store.dispatch(push({ pathname:'/', search: '' })); // Drops the params
});

检查控制台，您会看到查询参数（搜索）字段为空

Answer 4

查看我的路由器部分

 <div>
                <MainHeader />
                <Switch>
                    <Route exact path='/:pickupLocationName/:pickupDate/:pickupTime/:returnDate/:returnTime' render={(props) => <Quote  {...props} />} />
                    <Route path='/BookerInitial/:user/:id' component={BookerInitial} />
                    <Route path='/VehicleList' component={VehicleList} />
                    <Route path='/AdditionalCoverages' component={AdditionalCoverages} />
                    <Route path='/BookingDetails' component={BookingDetails} />
                    <Route path='/RenterInfo' component={RenterInfo} />
                </Switch>
                <Footer/>
            </div>

BookerInitial 页面中有下一个按钮

<button className="submit btn-skew-r btn-effect" id="nextBtn" style={{ backgroundColor: "#da1c36" }} onClick={this.handleSubmit}  >Next</button>

提交按钮方法

 handleSubmit = () => {
      this.props.history.push('./VehicleList');

    }

我也遇到了这个问题。 最后我找到了我改变的解决方案 this.props.history.push('/VehicleList'); 来自 this.props.history.push('./VehicleList'); 上面你可以看到点（。）是我的代码的问题。 所以不需要通过代码删除 URL 中的参数。 谢谢你