React TypeScript HoC - 将组件作为道具传递

Question

遵循本教程： https ://reacttraining.com/react-router/web/example/auth-workflow。

尝试重现代码：

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

在打字稿中：

import * as React from 'react';
import { Route, RouterProps } from 'react-router';

interface Props extends RouterProps {
  component: React.Component;
}

const PrivateRoute = ({ component: Component, ...rest }: Props) => {
  return (
    <Route
      {...rest}
      render={(props) => <Component {...props} />}
    />
  );
};

export default PrivateRoute;

但它总是会失败。 尝试了不同的变化。 我发布的是最新的一个。 得到：

在我看来，我必须为 Component 类型传递 Generic，但我不知道如何。

编辑：

到目前为止最接近的解决方案：

interface Props extends RouteProps {
  component: () => any;
}

const PrivateRoute = ({ component: Component, ...rest }: Props) => {
  return (
    <Route
      {...rest}
      render={(props) => <Component {...props} />}
    />
  );
};

接着：

<PrivateRoute component={Foo} path="/foo" />

Answer 1

你想传递一个组件构造函数，而不是一个组件实例：

import * as React from 'react';
import { Route, RouteProps } from 'react-router';

interface Props extends RouteProps {
    component: React.ComponentType;
}

const PrivateRoute = ({ component: Component, ...rest }: Props) => {
    return (
        <Route
            {...rest}
            render={(props) => <Component {...props} />}
        />
    );
};

export default PrivateRoute;

class Foo extends React.Component {

}
let r = <PrivateRoute component={Foo} path="/foo" />

编辑

更完整的解决方案应该是通用的，并使用RouteProps代替RouterProps ：

import * as React from 'react';
import { Route, RouteProps } from 'react-router';

type Props<P> =  RouteProps & P & {
    component: React.ComponentType<P>;
}

const PrivateRoute = function <P>(p: Props<P>) {
    // We can't use destructuring syntax, because : "Rest types may only be created from object types", so we do it manually.
    let rest = omit(p, "component");
    let Component = p.component;
    return (
        <Route
            {...rest}
            render={(props: P) => <p.component {...props} />}
        />
    );
};

// Helpers
type Diff<T extends string, U extends string> = ({[P in T]: P } & {[P in U]: never } & { [x: string]: never })[T];  
type Omit<T, K extends keyof T> = Pick<T, Diff<keyof T, K>>; 
function omit<T, TKey extends keyof T>(value:T, ... toRemove: TKey[]): Omit<T, TKey>{
    var result = Object.assign({}, value);
    for(let key of toRemove){
        delete result[key];
    }
    return result;
}


export default PrivateRoute;

class Foo extends React.Component<{ prop: number }>{

}
let r = <PrivateRoute component={Foo} path="/foo" prop={10} />

Answer 2

经过几个小时的调查，这是符合我要求的解决方案：

import * as React from 'react';
import { Route, RouteComponentProps, RouteProps } from 'react-router';

const PrivateRoute: React.SFC<RouteProps> =
  ({ component: Component, ...rest }) => {
    if (!Component) {
      return null;
    }
    return (
      <Route
        {...rest}
        render={(props: RouteComponentProps<{}>) => <Component {...props} />}
      />
    );
  };

export default PrivateRoute;

---

• 没有any ;
• 没有额外的复杂性；
• 构图保留；