AWS Lambda:如何从简单的java类调用lambda函数

我创建了简单的Lambda函数并将其上传到AWS Lambda.

import com.amazonaws.services.lambda.runtime.Context;
import com.amazonaws.services.lambda.runtime.RequestHandler;

public class Hello implements RequestHandler {

    @Override
    public String handleRequest(String input, Context context) {
         String output = "Bonjour, " + input + "!";
         return output;
    }

}

}

我想使用java类从其他项目调用此Lambda函数.我aws-java-sdk-lambda-1.10.22用来调用这个函数.但我无法成功.

这是我的InvokeLambda类,它是一个单独的项目.

import java.nio.ByteBuffer;
import java.nio.charset.Charset;

import org.apache.commons.logging.Log;
import org.apache.commons.logging.LogFactory;

import com.amazonaws.auth.AWSCredentials;
import com.amazonaws.auth.BasicAWSCredentials;
import com.amazonaws.regions.Region;
import com.amazonaws.regions.Regions;
import com.amazonaws.services.lambda.AWSLambdaClient;
import com.amazonaws.services.lambda.model.InvokeRequest;

public class InvokeLambda {
    private static final Log logger = LogFactory.getLog(InvokeLambda.class);
    private static final String awsAccessKeyId = "XXXXXX";
    private static final String awsSecretAccessKey = "YYYY";
    private static final String regionName = "us-west-2";
    private static final String functionName = "Hello";
    private static Region region;
    private static AWSCredentials credentials;
    private static AWSLambdaClient lambdaClient;

    /**
     * The entry point into the AWS lambda function.
     */
    public static void main(String... args) {
        credentials = new BasicAWSCredentials(awsAccessKeyId,
                awsSecretAccessKey);

        lambdaClient = (credentials == null) ? new AWSLambdaClient()
                : new AWSLambdaClient(credentials);
        //lambdaClient.configureRegion(Regions.US_WEST_2);
        region = Region.getRegion(Regions.fromName(regionName));
        lambdaClient.setRegion(region);

        try {
            InvokeRequest invokeRequest = new InvokeRequest();
            invokeRequest.setFunctionName(functionName);
            invokeRequest.setPayload("\" AWS Lambda\"");
            System.out.println(byteBufferToString(
                    lambdaClient.invoke(invokeRequest).getPayload(),
                    Charset.forName("UTF-8")));
        } catch (Exception e) {
            logger.error(e.getMessage());
            // System.out.println(e.getMessage());

        }
    }

    public static String byteBufferToString(ByteBuffer buffer, Charset charset) {
        byte[] bytes;
        if (buffer.hasArray()) {
            bytes = buffer.array();
        } else {
            bytes = new byte[buffer.remaining()];
            buffer.get(bytes);
        }
        return new String(bytes, charset);
    }
}

如何使用java调用lambda函数？

根据您的评论中的信息,您调用该函数的客户端代码很好.问题似乎与功能本身的配置有关.具体来说,AWS Lambda无法找到您指定的处理程序(com.aws.HelloLambda::handleRequest),因为它与处理程序类(Hello)的名称和包以及该类()中方法的名称不匹配handleRequest.

您可以通过AWS控制台更新功能处理程序名称.选择您的函数,然后选择配置选项卡,然后选择Handler属性.

您可能想要将其更改com.aws.HelloLambda::handleRequest为Hello::handleRequest.

在从Java客户端测试函数之前,您可以直接通过控制台对其进行测试,这将有助于确保正确配置该函数.