typescript - 基于输入参数的 TypeScript 函数返回类型

Question

我有几个不同的接口和对象，每个接口和对象都有一个type属性。假设这些是存储在 NoSQL 数据库中的对象。如何getItem根据输入参数创建具有确定性返回类型的泛型函数type？

interface Circle {
    type: "circle";
    radius: number;
}

interface Square {
    type: "square";
    length: number;
}

const shapes: (Circle | Square)[] = [
    { type: "circle", radius: 1 },
    { type: "circle", radius: 2 },
    { type: "square", length: 10 }];

function getItems(type: "circle" | "square") {
    return shapes.filter(s => s.type == type);
    // Think of this as items coming from a database
    // I'd like the return type of this function to be
    // deterministic based on the `type` value provided as a parameter. 
}

const circles = getItems("circle");
for (const circle of circles) {
    console.log(circle.radius);
                       ^^^^^^
}

类型“圆”上不存在属性“半径”| 正方形'。

标签： typescript

Answer 1

Conditional Types救援：

interface Circle {
    type: "circle";
    radius: number;
}

interface Square {
    type: "square";
    length: number;
}

type TypeName = "circle" | "square"; 

type ObjectType<T> = 
    T extends "circle" ? Circle :
    T extends "square" ? Square :
    never;

const shapes: (Circle | Square)[] = [
    { type: "circle", radius: 1 },
    { type: "circle", radius: 2 },
    { type: "square", length: 10 }];

function getItems<T extends TypeName>(type: T) : ObjectType<T>[]  {
    return shapes.filter(s => s.type == type) as ObjectType<T>[];
}

const circles = getItems("circle");
for (const circle of circles) {
    console.log(circle.radius);
}

感谢 Silvio 为我指明了正确的方向。