资产非空的声明合并

Question

我想对 @types/aws-lambda 进行调整，以表达我们的意图，即 pathParameters 不为空并且具有特定格式。

export interface APIGatewayProxyEventBase<TAuthorizerContext> {
    body: string | null;
    headers: APIGatewayProxyEventHeaders;
    multiValueHeaders: APIGatewayProxyEventMultiValueHeaders;
    httpMethod: string;
    isBase64Encoded: boolean;
    path: string;
    pathParameters: APIGatewayProxyEventPathParameters | null;
    // ...snip...
}
export interface APIGatewayProxyEventPathParameters {
    [name: string]: string | undefined;
}

在我们的应用程序代码中，我们可以表示 fooId 不为空

declare module "aws-lambda/trigger/api-gateway-proxy" {
  export interface APIGatewayProxyEventPathParameters {
    fooId: string;
  }
}

export const handler: APIGatewayProxyHandler = async (event) => {
  // This is still needed
  if (!event.pathParameters) {
    throw new Error("parameter is empty");
  }
  // Without declaration merging, fooId is string | undefined 
  // but is now string
  const { fooId } = event.pathParameters;

为了删除if (!event.pathParameters)，我写了这个：

declare module "aws-lambda/trigger/api-gateway-proxy" {
  export interface APIGatewayProxyEventBase<T> {
    pathParameters: APIGatewayProxyEventPathParameters;
  }
  export interface APIGatewayProxyEventPathParameters {
    fooId: string
  }
}

并给出以下错误。

error TS2428: All declarations of 'APIGatewayProxyEventBase' must have identical type parameters.
error TS2717: Subsequent property declarations must have the same type.  Property 'pathParameters' must be of type 'APIGatewayProxyEventPathParameters | null', but here has type 'APIGatewayProxyEventPathParameters'.

event 可以是 { pathParameters: { fooId: string }, ... } 类型而不是 { pathParameters?: { fooId: string }, ... } 吗？

Answer 1

您可以使用Omit utility type 来完成此操作。

interface LambdaEvent extends Omit<APIGatewayProxyEventV2, 'pathParameters'> {
  pathParameters: { fooId: string }
}

然后使用修改后的LambdaEvent 类型在您的处理程序中键入事件参数

export const handler: Handler<LambdaEvent, APIGatewayProxyResultV2> = (event) => { ... }

这在处理程序类型file 中有示例