Question

我有这样的功能：

const createRelationship = <T extends { id: string }>(
  includedItems: Array<DataResponse<T>>,
  categories: Array<{ id: string, type: string }>,
): Array<T> => {
  return categories.map(({ id }) =>
    includedItems.find((includedItem) => includedItem.id === id) as DataResponse<T>)
  .map(flatAttributes);
};

它采用通用类型T。第一个参数是DataResponse数组，其中：

interface DataResponse<T extends { id: string }, R = void> {
  id: string;
  attributes: Omit<T, 'id'>;
}

和：

type Omit<A extends object, K extends keyof A> = Pick<A, Exclude<keyof A, K>>;

如此

type Example = DataResponse<{ id: string, key: number }> // { id: string, attributes: { key: number } };

函数flatAttributes如下所示：

const flatAttributes = <T extends { id: string, attributes: any }>(item: T): { id: string } & T['attributes'] =>
  ({...item.attributes, ...item });

这是一种反向操作，例如：

flatAttributes{ id: string, attributes: { key: number } }) === { id: string, key: number }

现在为止：createRelationship的汇编：

Type '({ id: string; } & Pick<T, Exclude<keyof T, "id">>)[]' is not assignable to type 'T[]'.
  Type '{ id: string; } & Pick<T, Exclude<keyof T, "id">>' is not assignable to type 'T'.

Pick<T, Exclude<keyof T, "id">>'是Omit<T, 'key'>，而Omit<T, 'key'> & {key: string}应该是相同的T。

任何想法如何在不安全映射的情况下声明此函数？为什么TypeScript会表现得如此？

Playground

TypeScript：键入'{id：string; }＆Pick <t，排除<keyof =“” t，=“”“ id” =“” >>>>'不能分配给类型'T'

0 个答案: