执行"Hello World"时出错适用于Java中的AWS Lambda [英] Error executing "Hello World" for AWS Lambda in Java

问题描述

我写了我正在执行的Hello World Lambda，它是通过用于Eclipse的AWS工具包在AWS上上传来执行的.

I have written following Hello World Lambda which I am executing by uploading on AWS via AWS toolkit for Eclipse.

public class HelloWorldLambdaHandler implements RequestHandler<String, String> {
    public String handleRequest(String input, Context context) {
        System.out.println("Hello World! executed with input: " + input);
        return input;
    }
}

执行上述代码时出现以下错误.知道我在这里可能做错了什么吗? BTW Maven项目具有此处理程序，没有任何其他类，仅依赖项是aws-lambda-java-core版本1.1.0.

I am getting following error when executing above code. Any idea what I maybe doing wrong here? BTW Maven project which have this handler, doesn't have any other class and only dependency is aws-lambda-java-core version 1.1.0.

Skip uploading function code since no local change is found...
Invoking function...
==================== FUNCTION OUTPUT ====================
{"errorMessage":"An error occurred during JSON parsing","errorType":"java.lang.RuntimeException","stackTrace":[],"cause":{"errorMessage":"com.fasterxml.jackson.databind.JsonMappingException: Can not deserialize instance of java.lang.String out of START_OBJECT token\n at [Source: lambdainternal.util.NativeMemoryAsInputStream@2f7c7260; line: 1, column: 1]","errorType":"java.io.UncheckedIOException","stackTrace":[],"cause":{"errorMessage":"Can not deserialize instance of java.lang.String out of START_OBJECT token\n at [Source: lambdainternal.util.NativeMemoryAsInputStream@2f7c7260; line: 1, column: 1]","errorType":"com.fasterxml.jackson.databind.JsonMappingException","stackTrace":["com.fasterxml.jackson.databind.JsonMappingException.from(JsonMappingException.java:148)","com.fasterxml.jackson.databind.DeserializationContext.mappingException(DeserializationContext.java:835)","com.fasterxml.jackson.databind.deser.std.StringDeserializer.deserialize(StringDeserializer.java:59)","com.fasterxml.jackson.databind.deser.std.StringDeserializer.deserialize(StringDeserializer.java:12)","com.fasterxml.jackson.databind.ObjectReader._bindAndClose(ObjectReader.java:1441)","com.fasterxml.jackson.databind.ObjectReader.readValue(ObjectReader.java:1047)"]}}}

推荐答案

出于某种原因，Amazon无法将json反序列化为字符串.您会认为String会尽可能地与输入参数一样通用，但是对或错是不兼容的.

For some reason Amazon can't deserialize json to a String. You would think String would be as general as input parameter as you can get but rightly or wrongly it's not compatible.

要处理JSON，您可以使用Map或自定义POJO.

To handle JSON you can either use a Map or a custom POJO.

public class HelloWorldLambdaHandler {
    public String handleRequest(Map<String,Object> input, Context context) {
        System.out.println(input);
        return "Hello";
    }
}

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