Create own react route class in typescript

问题

I found this (reacttraining.com) site, which explains react-router with some examples. But I am not be able to do this with a typescript class. What I want to do is extend the Route class to build my own one. Right now I want to implement it in typescript for authentication as in the following example from the site.

const PrivateRoute = ({ component, ...rest }) => (
  <Route {...rest} render={props => (
    fakeAuth.isAuthenticated ? (
      React.createElement(component, props)
    ) : (
      <Redirect to={{
        pathname: '/login',
        state: { from: props.location }
      }}/>
    )
  )}/>
)

I searched a lot, but couldn't find a site that explains the function to implement and which typed properties to call to nested routes. An ES6 class will be also helpful, thank you.

回答1:

Here's my best shot so far, although there's still one any remaining :)

import * as React from "react"
import {Redirect, Route, RouteComponentProps, RouteProps} from "react-router-dom"

type RouteComponent = React.StatelessComponent<RouteComponentProps<{}>> | React.ComponentClass<any>

const AUTHENTICATED = false // TODO: implement authentication logic

export const PrivateRoute: React.StatelessComponent<RouteProps> = ({component, ...rest}) => {
  const renderFn = (Component?: RouteComponent) => (props: RouteProps) => {
    if (!Component) {
      return null
    }

    if (AUTHENTICATED) {
      return <Component {...props} />
    }

    const redirectProps = {
      to: {
        pathname: "/auth/sign-in",
        state: {from: props.location},
      },
    }

    return <Redirect {...redirectProps} />
  }

  return <Route {...rest} render={renderFn(component)} />
}

回答2:

Regarding Redux ...

Jacka's answer helped me alot, but i had a difficult time connecting the PrivateRoute component to redux. Furthermore i wanted to abstract the resulting Route component to work e.g. as a LoggedInRoute, NotLoggedInRoute or in general a Route which presents it's component if a condition is fulfilled or redirects to a specified location otherwise:

Note: Written with redux 4, react-router-dom 4 and typescript 2.9.

import * as H from 'history';
import * as React from 'react';
import { connect, MapStateToPropsParam } from 'react-redux';
import { Redirect, Route, RouteComponentProps, RouteProps } from 'react-router';

export interface ConditionalRouteProps extends RouteProps {
  routeCondition: boolean;
  redirectTo: H.LocationDescriptor;
}

export class ConditionalRoute extends React.Component<ConditionalRouteProps> {
  public render() {
    // Extract RouteProps without component property to rest.
    const { component: Component, routeCondition, redirectTo, ...rest } = this.props;
    return <Route {...rest} render={this.renderFn} />
  }

  private renderFn = (renderProps: RouteComponentProps<any>) => {
    if (this.props.routeCondition) {
      const { component: Component } = this.props; // JSX accepts only upprcase.
      if (!Component) {
        return null;
      }
      return <Component {...renderProps} />
    }

    return <Redirect to={this.props.redirectTo} />;
  };
}

export function connectConditionalRoute<S>(mapStateToProps: MapStateToPropsParam<ConditionalRouteProps, RouteProps, S>) {
  return connect<ConditionalRouteProps, {}, RouteProps, S>(mapStateToProps)(ConditionalRoute);
}

You can either use the ConditionalRoute component without connecting it and use your component's local state, e.g.:

interface RootState {
  loggedIn: boolean;
}

export class Root extends React.Component<RootProps, RootState> {
  /* skipped initialState and setState(...) calls */

  public render() {
    return (
      <Switch>
        <ConditionalRoute
          path="/todos"
          component={TodoPage}
          routeCondition={this.state.loggedIn}
          redirectTo="/login" />
        <ConditionalRoute
          path="/login"
          component={LoginPage}
          routeCondition={!this.state.loggedIn}
          redirectTo="/" />
        <Redirect to="/todos" />
      </Switch>
    );
  }
}

Or use the utility function connectConditionalRoute<S>(...) to use your redux store:

const loginRoute = '/login';
const todosRoute = '/todos';

const LoggedInRoute = connectConditionalRoute<RootState>(state => ({
  redirectTo: loginRoute,
  routeCondition: state.isLoggedIn,
}));

const NotLoggedInRoute = connectConditionalRoute<RootState>(state => ({
  redirectTo: todosRoute,
  routeCondition: !state.isLoggedIn
}));

const Root: React.SFC = () => (
  <Switch>
    <LoggedInRoute path="/todos" component={TodoPage} />
    <NotLoggedInRoute path="/login" component={LoginPage} />
    <Redirect to="/todos" />
  </Switch>
);

The behaviour in the provided sample: Unauthorized users visit /todos, get redirected to /login, authorized users visit /login, get redirected to /todos. Whenever the redux store's isLoggedIn changes, the connected components are updated and redirect the user automatically.

回答3:

I was looking for same thing. The question is old but maybe someone is still looking for it. Here is what I come up with (All types properly used from react-router 4):

interface PrivateRouteProps extends RouteProps {
  component: React.ComponentType<RouteComponentProps<any>> | React.ComponentType<any>
}
type RenderComponent = (props: RouteComponentProps<any>) => React.ReactNode;

export class PrivateRoute extends Route<PrivateRouteProps> {
  render () {
    const {component: Component, ...rest}: PrivateRouteProps = this.props;
    const renderComponent: RenderComponent = (props) => (
      AuthenticationService.isAuthenticated()
        ? <Component {...props} />
        : <Redirect to='/login' />
    );

    return (
      <Route {...rest} render={renderComponent} />
    );
  }
}

回答4:

here is my solution using "react-router-dom": "^4.4.0-beta.6" and "typescript": "3.2.2"

import React, { FunctionComponent } from "react";
import {
  Route, 
  Redirect,
  RouteProps, 
  RouteComponentProps
} from "react-router-dom";

interface PrivateRouteProps extends RouteProps {
  component:
    | React.ComponentType<RouteComponentProps<any>>
    | React.ComponentType<any>;
}

const PrivateRoute: FunctionComponent<PrivateRouteProps> = ({
  component: Component,
  ...rest
}) => {
  return (
    <Route
      {...rest}
      render={props =>
        true ? ( //put your authenticate logic here
          <Component {...props} />
        ) : (
          <Redirect
            to={{
              pathname: "/signin"
            }}
          />
        )
      }
    />
  );
};

export default PrivateRoute;

回答5:

You could use any.

const PrivateRoute = ({component: Component, ...rest }: any) => (
  <Route {...rest} render={PrivateRender(Component)} />
);

const PrivateRender = (Component: any) => {
  return (props: any) => {
    return <Component {...props}/>;
  };
};

来源：https://stackoverflow.com/questions/42309708/create-own-react-route-class-in-typescript