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反应路由器条件重定向

[英]React Router conditional redirect

I'm building a search UI using React, React Router and the awesome Reactivesearch library.我正在使用 React、React Router 和很棒的Reactivesearch库构建搜索 UI。 I'm trying to figure out how I can prevent users from simply navigating to mydomain.com/search , since that is my search results route.我试图弄清楚如何阻止用户简单地导航到mydomain.com/search ,因为那是我的搜索结果路线。

Ideally, if users tried to navigate to mydomain.com/search, I will use RR Redirect component to redirect to the home page.理想情况下,如果用户尝试导航到 mydomain.com/search,我将使用 RR Redirect 组件重定向到主页。

I'm using "/search" for the route that the Route component in RR(v5) to render the search results page and can't quite figure out how to use something like /search?q=${value} to render the page?我正在为 RR(v5) 中的 Route 组件呈现搜索结果页面的路线使用"/search" ,并且无法完全弄清楚如何使用/search?q=${value}之类的东西来呈现页?

As a preface I do have this in the render block (I'm using class based component for search results)作为前言,我在渲染块中有这个(我使用基于类的组件来搜索结果)

let value = JSON.stringify(queryString.parse(location.search));
if (this.value === '' || null) {
       return (
         <Redirect to="/" />
       );
     }

However, its not working... I can still go to my address bar and type in mydomain.com/search and the page renders.但是,它不起作用...我仍然可以转到地址栏并输入 mydomain.com/search 并呈现页面。

Here is an example in my SearchResults.tsx:这是我的 SearchResults.tsx 中的一个示例:

<Route path = "/search" render={() => (
   <ReactiveList
    ...
    ...
    />
   />

I'm trying to get to我试图到达

<Route path = `/search?q="${value}"` render={() => (
   <ReactiveList
    ...
    ...
    />
   />

Update Docs on ReactiveList Example from docs:从文档更新有关ReactiveList示例的文档:

<ReactiveList
    componentId="SearchResult"
    dataField="ratings"
    pagination={false}
    paginationAt="bottom"
    pages={5}
    sortBy="desc"
    size={10}
    loader="Loading Results.."
    showResultStats={true}
    renderItem={res => <div>{res.title}</div>}
    renderResultStats={function(stats) {
        return `Showing ${stats.displayedResults} of total ${stats.numberOfResults} in ${
            stats.time
        } ms`;
    }}
    react={{
        and: ['CitySensor', 'SearchSensor'],
    }}
/>

How can I prevent users from simply navigating to mydomain.com/search ??如何防止用户简单地导航到 mydomain.com/search ?

If you want to conditionally render the ReactiveList component based on if there's a truthy q queryString parameter then you can use either a wrapper component, a layout route, or a Higher Order Component (HOC) to read the queryString and handle the redirection logic.如果您想根据是否有真值q queryString 参数有条件地呈现ReactiveList组件,那么您可以使用包装器组件、布局路由或高阶组件 (HOC) 来读取 queryString 并处理重定向逻辑。

react-router-dom@6

Using a Wrapper使用包装器

import { Navigate, useSearchParams } from 'react-router-dom';

const QuaryParametersWrapper = ({ children, parameters = [] }) => {
  const [searchParams] = useSearchParams();

  const hasParameter = parameters.some(
    (parameter) => !!searchParams.get(parameter)
  );

  return hasParameter ? children : <Navigate to="/" replace />;
};

... ...

<Route
  path="/search"
  element={(
    <ReactiveListWrapper parameters={["q"]}>
      <ReactiveList
        ...
        ...
      />
    </ReactiveListWrapper>
  )}
/>

Using a custom HOC使用自定义 HOC

import { Navigate, useSearchParams } from 'react-router-dom';

const withQueryParameters = (...parameters) => (Component) => (props) => {
  const [searchParams] = useSearchParams();

  const hasParameter = parameters.some(
    (parameter) => !!searchParams.get(parameter)
  );

  return hasParameter ? <Component {...props} /> : <Navigate to="/" replace />;
};

... ...

export default withQueryParameters("q")(ReactiveList);

... ...

import ReactiveListWrapper from '../path/to/ReactiveList';

...

<Route path="/search" element={<ReactiveListWrapper />} />

Using a layout route使用布局路线

import { Navigate, Outlet, useSearchParams } from 'react-router-dom';

const QuaryParametersLayout = ({ parameters = [] }) => {
  const [searchParams] = useSearchParams();

  const hasParameter = parameters.some(
    (parameter) => !!searchParams.get(parameter)
  );

  return hasParameter ? <Outlet /> : <Navigate to="/" replace />;
};

... ...

<Route element={<QuaryParametersLayout parameters={["q"]} />}>
  <Route path="/search" element={<ReactiveList />} />
</Route>

Demo演示

编辑反应路由器条件重定向

react-router-dom@5

The useSearchParams hook doesn't exist in v5 so you can create your own. v5 中不存在useSearchParams挂钩,因此您可以创建自己的挂钩。

import { useMemo } from 'react';
import { useLocation } from 'react-router-dom';

const useSearchParams = () => {
  const { search } = useLocation();
  const searchParams = useMemo(() => new URLSearchParams(search), [search]);
  return [searchParams];
};

Using a Wrapper使用包装器

import useSearchParams from '../path/to/useSearchParams';

const QuaryParametersWrapper = ({ children, parameters = [] }) => {
  const [searchParams] = useSearchParams();

  const hasParameter = parameters.some(
    (parameter) => !!searchParams.get(parameter)
  );

  return hasParameter ? children : <Redirect to="/" />;
};

... ...

<Route
  path="/search1"
  render={(routeProps) => (
    <QuaryParametersWrapper parameters={["q"]}>
      <ReactiveList {...routeProps} />
    </QuaryParametersWrapper>
  )}
/>

Using a custom HOC使用自定义 HOC

import { Redirect } from 'react-router-dom';
import useSearchParams from '../path/to/useSearchParams';

const withQueryParameters = (...parameters) => (Component) => (props) => {
  const [searchParams] = useSearchParams();

  const hasParameter = parameters.some(
    (parameter) => !!searchParams.get(parameter)
  );

  return hasParameter ? <Component {...props} /> : <Redirect to="/" />;
};

... ...

export default withQueryParameters("q")(ReactiveList);

... ...

import ReactiveListWrapper from '../path/to/ReactiveList';

...

<Route path="/search2" component={QueryReactiveList} />

Demo演示

编辑 react-router-conditional-redirect (RRDv5)

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