React TypeScript HoC-将Component作为prop [英] React TypeScript HoC - passing Component as the prop
本文介绍了React TypeScript HoC-将Component作为prop的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
按照本教程操作: https://reacttraining.com/react-router/网络/示例/身份验证工作流.
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TypeScript中:
In TypeScript:
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;
但是它总是会失败.尝试了不同的变化.我发布的是最新的一个.正在获取:
But it would always fail. Tried different variations. The one I've posted the most recent one. Getting:
在我看来,我必须将Generic作为Component类型传递,但我不知道如何.
It seems to me that I have to pass Generic for the Component type, but I don't know how.
到目前为止最接近的解决方案:
The closest solution so far:
interface Props extends RouteProps {
component: () => any;
}
const PrivateRoute = ({ component: Component, ...rest }: Props) => {
return (
<Route
{...rest}
render={(props) => <Component {...props} />}
/>
);
};
然后:
<PrivateRoute component={Foo} path="/foo" />
推荐答案
您要传递组件构造函数,而不是组件实例:
You want to pass a component constructor, not a component instance:
import * as React from 'react';
import { Route, RouteProps } from 'react-router';
interface Props extends RouteProps {
component: new (props: any) => React.Component;
}
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" />
修改
一个更完整的解决方案应该是通用的,并使用RouteProps
代替RouterProps
:
A more complete solution should be generic and use RouteProps
instead RouterProps
:
import * as React from 'react';
import { Route, RouteProps } from 'react-router';
type Props<P> = RouteProps & P & {
component: new (props: P) => React.Component<P>;
}
const PrivateRoute = function <P>(p: Props<P>) {
// We can't use destructureing 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} />
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