考察如下场景:
- 一个自定义的下拉选择框有个
type 属性包含两种可能的值 "native" | "simulate"
- 当
type 为 simulate 时还希望传递一个 appearence 控制其样式
- 当
type 为 native 时则不希望传递 appearence 属性
即 appearence 属性是否通过 TypeScript 的类型检查依赖于 type 的值,请问组件的属性类型如何定义。
一开始会以为这里需要借助泛型等手段来构造一个复杂类型,其实大可不必。因为后来一想不防用 Union Types 试试,实践后证实,事情其实没想的那么复杂。
类型定义:
type SelectProps =
| {
type: "native";
}
| {
type: "simulate";
appearance: "default" | "link" | "button";
};
使用:
function CustomSelect(props: SelectProps) {
return <div>...</div>;
}
function App() {
return (
<>
{/* ✅ */}
<CustomSelect type="native" />
{/* ❌ Type '{ type: "native"; appearance: string; }' is not assignable to type 'IntrinsicAttributes & SelectProps'.
Property 'appearance' does not exist on type 'IntrinsicAttributes & { type: "native"; }'. */}
<CustomSelect type="native" appearance="button" />
{/* ❌ Type '{ type: "simulate"; }' is not assignable to type 'IntrinsicAttributes & SelectProps'.
Property 'appearance' is missing in type '{ type: "simulate"; }' but required in type '{ type: "simulate"; appearance: "default" | "link" | "button"; }'. */}
<CustomSelect type="simulate" />
{/* ✅ */}
<CustomSelect type="simulate" appearance="button" />
</>
);
}
|
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