React TypeScript：路由位置和子属性的正确类型

Question

I have a router that passes props of location and children, but I don't know what the correct types are for these props.

This is the router using react-router-dom...

import React, { useReducer } from 'react';
import { BrowserRouter, Route, Switch } from 'react-router-dom';
import { globalContext, setGlobalContext } from './components/context';
import { Layout } from './components/layout';
import { defaultState, globalReducer } from './components/reducer';
import { Home } from './routes/home';
import { NotFound } from './routes/notFound';

const Router: React.FC = () => {
  const [global, setGlobal] = useReducer(globalReducer, defaultState);
  return (
    <setGlobalContext.Provider value={{ setGlobal }}>
      <globalContext.Provider value={{ global }}>
        <BrowserRouter>
          <Route
            render={({ location }) => (
              <Layout location={location}>
                <Switch location={location}>
                  <Route exact path='/' component={Home} />
                  <Route component={NotFound} />
                </Switch>
              </Layout>
            )}
          />
        </BrowserRouter>
      </globalContext.Provider>
    </setGlobalContext.Provider>
  );
};

export { Router };

and inside layout component, I have an interface for the location and children, but because I don't know the correct type for these I end up using any, my linter then throws a fit.

import React from 'react';

interface PropsInterface {
  children: any;
  location: any;
}

const Layout: React.FC<PropsInterface> = (props: PropsInterface) => {
  return (
    <div>
      <p>{props.location}</p>
      <p>{props.children}</p>
    </div>
  );
};

export { Layout };

The error I get is Type declaration of 'any' loses type-safety. Consider replacing it with a more precise type. (no-any)tslint(1)

Could I get location from import import { useLocation } from 'react-router-dom';

Answer 1

The interface for Route that you're looking for can be found in the @types/react-router definition.

After installing that in your project, you should be able to silence that TS error with an interface like this:

// Layout.tsx

import { RouteProps } from "react-router";

interface ILayoutProps {
  location: RouteProps["location"];
  children: RouteProps["children"];
}

const Layout: React.FC<ILayoutProps> = (props: ILayoutProps) => {
  return <div>{props.children}</div>;
};

Keep in mind, location is an object, so React isn't going to let you render that as you were previously.

Here's a link to a working sample: https://codesandbox.io/s/zealous-snyder-yw9dg

---

Another option is to use a Partial interface, but that will make all the ILayoutProps optional:

interface ILayoutProps extends Partial<RouteProps> {}

Answer 2

You can also find the interface for these types using the RouteComponentProps exposed by react-router-dom . Try:

import { RouteComponentProps } from "react-router-dom";

and map that to your Component like so:

// Layout.tsx

import { RouteComponentProps } from "react-router-dom";

const Layout: React.FC<RouteComponentProps> = (props) => {
  return <div>{props.children}</div>;
};