如何从给定的对象数组动态推断/映射类型？

Question

我有一个类似这样的对象数组：

const options = [
  { name: "diameter", type: "number", value: 10, min: 0, max: 10 },
  { name: "height", type: "number", value: 2, min: 0, max: 8 },
  { name: "thickness", type: "number", value: 2, min: 0, max: 8 },
  { name: "text", type: "string", value: "Hello World" },
];

我想动态获取类型，所以有这样的东西；

type OptionsType = {
  diameter: { value: number; min: number; max: number };
  height: { value: number; min: number; max: number };
  thickness: { value: number; min: number; max: number };
  text: { value: string };
};

我尝试四处寻找，询问chatGPT / perplexity和supermaven，但没有找到解决方案。

关闭人工智能的解决方案尝试是这样的：

// Helper type to determine the correct parameter type based on 'type' property
type ParamType<T extends { type: string }> = T["type"] extends "number"
  ? { value: number; min: number; max: number }
  : { value: string };

// Mapped type to transform the options array into the desired type definition
type OptionsParams<T extends readonly { name: string; type: string }[]> = {
  [K in T[number] as K["name"]]: ParamType<Extract<T[number], { name: K }>>;
};

// Generate the type dynamically
type optionsParams = OptionsParams<typeof options>;

但是这个 optionsParams 的返回类型是：

type optionsParams = {
    [x: string]: {
        value: number;
        min: number;
        max: number;
    };
}

所以它仅适用于（某种）数字值。

有什么建议如何解决这个问题，以及如何从 TypeScript 中给定的对象数组动态生成/映射类型吗？

Answer 1

如果你可以将

name

断言为 const，那么这将有效：

它从记录值类型中省略了

name

和

type

，但关键是

name

的const断言。

const options = [
  { name: "diameter" as const, type: "number", value: 10, min: 0, max: 10 },
  { name: "height" as const, type: "number", value: 2, min: 0, max: 8 },
  { name: "thickness" as const, type: "number", value: 2, min: 0, max: 8 },
  { name: "text" as const, type: "string", value: "Hello World" },
];

// Mapped type to transform the options array into the desired type definition
type OptionsParams<T extends readonly { name: string; type: string }[]> = Partial<{
  [K in T[number] as K["name"]]: Omit<K, "name" | "type">;
}>;

// Generate the type dynamically
type optionsParams = OptionsParams<typeof options>;

游乐场