typescript 的类型处理

This topic created in 855 days ago, the information mentioned may be changed or developed.

1 、interface

interface TypeD{
    subTypeA: string;
    subTypeB: number;
}
interface TypeProp{
    typeA: string;
    typeB: number;
    typeC: boolean;
    typeD: TypeD;
}

2 、data

const data1: TypeProp = {
    typeA: 'yes',
    typeB: 1,
    typeC: true,
    typeD: {
    	subTypeA: 'no',
        subTypeB: 0,
    }
}

3 、function

const dosomething = (data: keyof TypeProp):[?] => {
	return data1[data]
}

请问 [?] 处我需要怎么处理(或者我需要如何定义 dosomething)，才能表示返回的是 data1[data]的类型

我想要的

const b = dosomething('typeB') // b 是 number 类型
const c = dosomething('typeC') // c 是 boolean 类型
const d = dosomething('typeD') // d 是 TypeD 类型

8 replies • 2024-01-19 20:32:11 +08:00

lzgshsj

Jan 19, 2024

const dosomething = <T extends keyof TypeProp>(data: T): TypeProp[T] => {
return data1[data];
}

Gaoti

Jan 19, 2024

```typescript
function dosomething<K extends keyof TypeProp>(data: K): TypeProp[K] {
return data1[data];
}
```

善用泛型

yxcoder

Jan 19, 2024

@Gaoti
@lzgshsj

请问一下，如果 dosomething 里面涉及到数组应该怎么处理的

yxcoder

Jan 19, 2024

@yxcoder

譬如我希望
const dosomething = (data: keyof TypeProp | keyof TypeProp[]) => { ... }

yxcoder

Jan 19, 2024

@yxcoder const dosomething = (data: keyof TypeProp | (keyof TypeProp)[]) => { ... }

MinonHeart

Jan 19, 2024

可能类似这样吧，假设都返回对象形式
get(keys: keyof T): Pick<T, keyof T>;
get<U extends keyof T, K extends U[]>(keys: K): Pick<T, K[number]>;

Gaoti

Jan 19, 2024

@yxcoder #5

``` typescript
function dosomething<K extends keyof TypeProp>(data: K): TypeProp[K];

function dosomething<K extends keyof TypeProp>(
data: K[],
): { [P in K]: TypeProp[P] }[];

function dosomething<K extends keyof TypeProp>(
data: K | K[],
): TypeProp[K] | TypeProp[K][] {
if (Array.isArray(data)) {
return data.map((key) => data1[key]);
} else {
return data1[data];
}
}
```

bgm004

Jan 19, 2024 via Android

类型体操里有这种。