如何从函数返回单个类型

Question

我正在尝试在Typescript(React(中构建一个辅助函数。我定义了一个基于响应数据返回object或object[]的函数。

现在，当我使用该函数时，返回typeT | T[]，这需要根据数据进行T或T[]。

我的助手函数

import { useQuery } from '@apollo/react-hooks';
import { flatMap } from 'lodash';
export default function QueryHelper<T> (document: any, variables?: {}) {
const { data: responseData, loading, error } = useQuery(document, { variables });
const object = (): T => responseData;
const array = (): T[] => flatMap(responseData);
let data;
if (flatMap(responseData).length === 1) {
data = object();
} else {
data = array();
}
return { data, loading, error };
}

调用函数

const objects = QueryHelper<Object>(multipleObjectsDocument); 
const object = QueryHelper<Object>(singleObjectDocument, { id });

返回类型

const Object: {
data: Object| Object[]; // This needs to be 1 type
loading: boolean;
error: ApolloError | undefined;
}

主要思想是我可以打电话例如;

const name = object.data.name';
const listOfName = objects.data.map(obj => obj.name);

现在我收到以下错误Property 'map' does not exist on type 'Object | Object[]'.

我还尝试根据 if 语句有条件地返回不同的变量，但这会返回;

const Object: {
object: Object;
loading: boolean;
error: ApolloError | undefined;
} | {
array: Object[];
loading: boolean;
error: ApolloError | undefined;
}

Answer 1

刚刚修复了它。

export default function QueryHelper<T> (document: any, variables?: {}): queryResponse<T> {
const { data: responseData, loading, error } = useQuery(document, { variables });
let data = responseData ?? [];
const array = flatMap(data);
if (!loading && array.length === 1) {
data = array[0];
} else {
data = array;
}
return { data, loading, error };
}

调用

const objects = QueryHelper<Object[]>(MultipleObjectsDocument).data; // returns type Object[]
const object = QueryHelper<Object>(ObjectDocument, { id }).data; // returns type Object