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TypeScript 道具的条件类型

[英]TypeScript conditional types for props

I have a component that can be in two different states, editing mode and viewing mode.我有一个组件可以处于两种不同的状态,编辑模式和查看模式。 To accommodate for that I have a union type to make this work, but I still get TypeScript complaining in the args of my component:为了适应这一点,我有一个联合类型来完成这项工作,但我仍然得到 TypeScript 在我的组件的 args 中抱怨:

type View = {
    isEditing: false;
    viewHandler: () => void;
    someOtherProp: any // this is unique to view mode
};

type Edit = {
    isEditing: true;
    editHandler: (id: string) => void;
};

export type MyProps = View | Edit;

It is then complaining here:然后在这里抱怨:

const Item: React.FC<MyProps> = ({
    isEditing = false,
    editHandler,
    ~~~~~~~~~~~
}) => {

Property 'editHandler' does not exist on type 'PropsWithChildren<MyProps>'.

What am I missing?我错过了什么?

I think it should also be present in your View type like this:我认为它也应该出现在您的 View 类型中,如下所示:

type View = {
  isEditing: false;
  editHandler?:(id: string) => void;
};

type Edit = {
  isEditing?: true;
  editHandler: (id: string) => void;
};

So the solution I used now is adding it and typing it as never :所以我现在使用的解决方案是添加它并将其输入为never

type View = {
    isEditing: false;
    viewHandler: () => void;
    editHandler?: never;
};

type Edit = {
    isEditing: true;
    editHandler: (id: string) => void;
    viewHandler?: never;
};

export type MyProps = View | Edit;

And at the point where the event handler gets invoked or passed down, I use a !在事件处理程序被调用或传递的地方,我使用! to assure TypeScript that this cannot be undefined.向 TypeScript 保证这不能未定义。 Kudos to Nadia who pointed me in the right direction in a comment above!感谢 Nadia 在上面的评论中为我指明了正确的方向!

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