我正在使用来自 React Router v4 的 Route 组件的渲染 Prop ,使用 Typescript 和 React 实现经过身份验证的路由。
路线:
import React from 'react';
import { Switch, Route } from 'react-router-dom';
import { ROUTES } from 'utils/constants';
import HomePage from 'components/pages/Home';
import GuestLogin from 'components/pages/GuestLogin';
import ProfilePage from 'components/pages/Profile';
import NotFoundPage from 'components/pages/NotFound';
import ResetPassword from 'components/pages/ResetPassword';
import SetPassword from 'components/pages/SetPassword';
import LoginContainer from 'containers/Login';
import PrivateRoute from './PrivateRoute';
const Routes: React.FunctionComponent = () => (
<Switch>
<Route path={ROUTES.LOGIN} component={LoginContainer} exact></Route>
<PrivateRoute
path={ROUTES.HOME}
component={HomePage}
></PrivateRoute>
<Route path={ROUTES.GUEST_LOGIN} component={GuestLogin}></Route>
<Route path={ROUTES.RESET_PASSWORD} component={ResetPassword}></Route>
<Route path={ROUTES.SET_PASSWORD} component={SetPassword}></Route>
<Route path={ROUTES.PROFILE} component={ProfilePage}></Route>
<Route component={NotFoundPage}></Route>
</Switch>
);
export default Routes;
私有(private)路线:
import React from 'react';
import { useAppContext } from 'containers/App/AppContext';
import { RouteProps, Route, Redirect } from 'react-router-dom';
import { ROUTES } from 'utils/constants';
const PrivateRoute: React.FunctionComponent<RouteProps> = ({
component: Component,
...routeProps
}) => {
const { isSignedIn } = useAppContext();
const ComponentToRender = Component as React.ElementType;
return (
<Route
{...routeProps}
render={(props) =>
isSignedIn ? (
<ComponentToRender {...props} />
) : (
<Redirect to={ROUTES.LOGIN} />
)
}
/>
);
};
export default PrivateRoute;
问题是我想调用 Prop 上的组件集,但是,每次我尝试这样做时,Typescript 都会抛出以下错误。
JSX element type 'Component' does not have any construct or call signatures. TS2604
原因似乎是 Route 的组件类型不是 Typescript 所期望的类型,如此处解释:https://github.com/microsoft/TypeScript/issues/28631 ,因此我刚刚创建了一个具有新类型 (ComponentToRender) 的副本。
有没有更好的方法来实现这个?也许会覆盖 RouteProps 组件元素?
谢谢!
最佳答案
终于明白错误并解决了!关键是组件属性和渲染函数在类型方面的处理方式不同。
通过使用 RouteProps
,Typescript 编译器实际上将组件属性设置为 ComponentType(根据 RouterProps 为“组件”指定的类型),它不能用于渲染函数,如索引中所示。 d.ts 文件。
export interface RouteProps {
location?: H.Location;
component?: React.ComponentType<RouteComponentProps<any>> | React.ComponentType<any>;
render?: (props: RouteComponentProps<any>) => React.ReactNode;
children?: ((props: RouteChildrenProps<any>) => React.ReactNode) | React.ReactNode;
path?: string | string[];
exact?: boolean;
sensitive?: boolean;
strict?: boolean;
}
只要将渲染函数移动到路由文件中而不使用 PrivateRoute 组件,这种行为就会变得明显。下面的代码按预期工作,因为编译器正确推断类型 (React.ReactNode)。
<Route
path={ROUTES.POINTS_HISTORY}
render={(props) =>
isSignedIn ? <PointsTransactions /> : <Redirect to={ROUTES.LOGIN} />
}
></Route>
因此,为了解决这个问题,我只是创建了一个新类型,其中只有我的用例所需的参数。
import React from 'react';
import { useAppContext } from 'containers/App/AppContext';
import { RouteProps, Route, Redirect } from 'react-router-dom';
import { ROUTES } from 'utils/constants';
type PrivateRouteProps = {
path: RouteProps['path'];
component: React.ElementType;
};
const PrivateRoute: React.FunctionComponent<PrivateRouteProps> = ({
component: Component,
...routeProps
}) => {
const { isSignedIn } = useAppContext();
return (
<Route
{...routeProps}
render={(props) =>
isSignedIn ? <Component /> : <Redirect to={ROUTES.LOGIN} />
}
/>
);
};
export default PrivateRoute;
关于reactjs - 使用 Typescript 和 React 在私有(private)路由中传递组件 Prop ,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/63463502/