Amazon LambdaJavaSDK，不同意服务名称

Question

我试图通过AWSJavaSDK调用AWS Lambda代码，但它产生了一个奇怪的错误，我只在谷歌上发现了一个缺陷，应该是不久前修复的。

The service name : "BackplaneExecutionService" identified from the payload does not agree with the service name : "AWSLambda" specified in the transport header (Service: AWSLambda; Status Code: 403; Error Code: AccessDeniedException; Request ID: c40b5e2e-807b-11e5-88d6-6f0496ed99c7)
com.amazonaws.AmazonServiceException: The service name : "BackplaneExecutionService" identified from the payload does not agree with the service name : "AWSLambda" specified in the transport header (Service: AWSLambda; Status Code: 403; Error Code: AccessDeniedException; Request ID: c40b5e2e-807b-11e5-88d6-6f0496ed99c7)
    at com.amazonaws.http.AmazonHttpClient.handleErrorResponse(AmazonHttpClient.java:710)
    at com.amazonaws.http.AmazonHttpClient.executeHelper(AmazonHttpClient.java:385)
    at com.amazonaws.http.AmazonHttpClient.execute(AmazonHttpClient.java:196)
    at com.amazonaws.services.lambda.AWSLambdaClient.invoke(AWSLambdaClient.java:387)
    at com.amazonaws.services.lambda.AWSLambdaClient.invoke(AWSLambdaClient.java:322)
    at .....

我的调用代码如下所示：

    BasicAWSCredentials androidClientCredentials = new BasicAWSCredentials(
            "cccccccc",
            "ccc+cc/cc"
    );
    AWSLambdaClient lambdaClient = new AWSLambdaClient(androidClientCredentials);
    lambdaClient.setRegion(Region.getRegion(Regions.US_EAST_1));
    lambdaClient.setEndpoint("https://cc.execute-api.us-east-1.amazonaws.com/prod/asdf");
    lambdaClient.setServiceNameIntern("execute-api");
    InvokeRequest request = new InvokeRequest();
    request.setInvocationType(InvocationType.RequestResponse);
    request.setFunctionName("arn:aws:lambda:us-east-1:1234:asdf:$LATEST");
    Map<String,String> payload = new HashMap<>(4);
    payload.put("username",U);
    payload.put("password", P);
    request.setPayload(ByteBuffer.wrap(new JSONObject(payload).toString().getBytes()));
    InvokeResult result = lambdaClient.invoke(request);

有什么想法或方向吗？

Answer 1

感谢@jarmod，此错误是由于我对如何使Lambda函数可访问感到困惑。从我需要创建“APIendpoint”的UI来看。在阅读了更多关于通过AWS创建API的信息后，这似乎是一个用于创建自定义endpointURL的特殊功能（我还没有完全理解该功能）。

我了解到您不需要“APIendpoint”，并且我能够删除setEndpoint和setServiceNameInter方法调用