如何使用模板文字类型 [REUPLOAD] 从接口获取特定键？ [复制]

Question

我有以下代码为给定的数据库表接收possibleColumns 的字符串数组。然后它使用传入的source 并尝试将其与包含{source}_id 的列匹配。

例如possibleColumns 可以包含 ["id", "name", "first_id", "second_id"] 并且源是 first 然后检查任何名为 first_id 的列并返回它。

逻辑很好，但返回的类型不正确，目前它只返回可能的值作为该类型中的任何列，但我想使用模板文字类型只返回包含<value>_id 的任何值。正如在 cmets 中看到的那样，我正在尝试使用 Extract 类型但没有运气。

/**
 * Takes a list of columns for a specific table and a source and checks for any columns that
 * match `${source}_id`.
 * @template T The desired output type
 * @param possibleColumns string array of columns of type T
 * @param source The source to convert in question
 * @returns a key of one of the columns of T or undefined if not matched
 */
export function getProviderIdColumn<T>(possibleColumns: (keyof T)[], source: string) {
    // Attempt to extract only keys that contain _id at the end
    // as Extract<keyof T extends `${infer _}_id` ? keyof T : never, T>
    const providerIdColumn = `${source}_id` as keyof T;
    if (possibleColumns.includes(providerIdColumn)) {
        return providerIdColumn;
    }

电流输出

"id" | "name" | "first_id" | "second_id"

期望的输出

"first_id" | "second_id"

我对打字稿的了解不是很好，所以请忽略任何滥用的术语。

最小的工作示例

export interface Model {
    id: number,
    name: string | null,
    third_id: string | null,
    source: string,
    first_id: string | null,
    second_id: string | null,
}

/**
 * Takes a list of columns for a specific table and a source and checks for any columns that
 * match `${source}_id`.
 * @template T The desired output type
 * @param possibleColumns string array of columns of type T
 * @param source The source to convert in question
 * @returns a key of one of the columns of T or undefined if not matched
 */
export function getProviderIdColumn<T>(possibleColumns: (keyof T)[], source: string) {
    // Attempt to extract only keys that contain _id at the end
    // as Extract<keyof T extends `${infer _}_id` ? keyof T : never, T>
    const providerIdColumn = `${source}_id` as keyof T;
    if (possibleColumns.includes(providerIdColumn)) {
        return providerIdColumn;
    }
    return undefined;

}

// A function here returns a string array of the keys in Model.
const columnInfo: (keyof Model)[] = ["id", "name", "first_id", "source", "second_id", "third_id"];

const source = "first";

// Returned values here are fine but I want to get the desired output.
const providerIdColumn = getProviderIdColumn<Model>(columnInfo, source);

Playground link

Answer 1

T.J. Crowder的原始答案

重新上传以匿名信息，因为我无法编辑答案。

我们可以通过使用 mapped type 来提取 _id 属性名称来做到这一点：

/**
 * Gets a union of the literal types in `T` that end with `_id`.
 */
type IdKeys<T extends any[]> = keyof {
    [Key in T[number] as Key extends `${string}_id` ? Key : never]: null;
};

然后在函数中使用它：

/**
 * Takes a list of columns for a specific table and a source and checks for any columns that
 * match `${source}_id`.
 * @template ObjectType The desired output type
 * @param possibleColumns string array of columns of type T
 * @param source The source to convert in question
 * @returns a key of one of the columns of T or undefined if not matched
 */
export function getProviderIdColumn<ObjectType>(
    possibleColumns: (keyof ObjectType)[],
    source: string
): IdKeys<(keyof ObjectType)[]> | undefined {
    // Sadly, we have to have a type assertion, since `source` is type `string`
    const providerIdColumn = `${source}_id` as IdKeys<(keyof ObjectType)[]>;
    if (possibleColumns.includes(providerIdColumn)) {
        return providerIdColumn;
    }
    return undefined;
}

结果是在您的示例调用中：

const providerIdColumn = getProviderIdColumn<Model>(columnInfo, source);

...providerIdColumn 的类型是 "boardex_id" | "duedil_id" | "zoom_info_id" | undefined。

Playground example