使用Flow和redux-navigation验证通用的React组件属性

时间:2018-08-09 14:14:58

标签: reactjs react-native react-navigation flowtype

我试图在分派新的屏幕操作时验证React组件是否正在接收正确的属性。

路线可能看起来像这样。

class MyCompAB extends React.Component<{a: number, b: string}> {}
class MyCompCD extends React.Component<{c: number, d: string}> {}
const myRoutes = {'/ab': MyCompAB, '/cd': MyCompCD};

我需要的是这样的东西。我只是还没有找到使openScreen验证道具是否适用于路线[名称]的方法。

type Screen<P> = ComponentType<P>;
type Routes<P> = {[key: string]: Screen<P>};

function openScreen<P, R: Routes<mixed>, K: $Keys<R>>(routes: R, name: K, props: P): void {
  // This needs to validate that props are valid for the component in routes[name].
  // This won't do <Component {...props} />. Instead redux dispatch magic happens here.
  type Pepe = $Call<R<P>, K>;
}

// $ExpectError. Should fail: a and b missing.
openScreen(myRoutes, '/ab', {});
// $ExpectError. Should fail: b has wrong type.
openScreen(myRoutes, '/ab', {a: 1, b: 1});
// Should work.
openScreen(myRoutes, '/ab', {a: 1, b: "1"});

// $ExpectError. Should fail: c and d missing.
openScreen(myRoutes, '/cd', {});
// $ExpectError. Should fail: d has wrong type.
openScreen(myRoutes, '/cd', {c: 1, d: 1});
// Should work.
openScreen(myRoutes, '/cd', {c: 1, d: "1"});

我如何进行这项工作?

1 个答案:

答案 0 :(得分:0)

您可以将function overloading用于次优解决方案,例如:

declare function openScreen<P: $PropertyType<MyCompAB, 'props'>, R: Routes<mixed>, K: '/ab'>(routes: R, name: K, props: P): void {

}

declare function openScreen<P: $PropertyType<MyCompCD, 'props'>, R: Routes<mixed>, K: '/bc'>(routes: R, name: K, props: P): void {

}