java - AWS Lambda RequestHandler 用于无效输出

Question

考虑一个用 Java 编写的简单 Lambda:

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

public class Hello implements RequestHandler<Integer, String>{
    public String handleRequest(int myCount, Context context) {
        return String.valueOf(myCount);
    }
}

处理程序接口(interface)定义为 RequestHandler<InputType, OutputType> ，但是当我的 Lambda 对事件使用react并且只是产生一些副作用时，输出类型是不必要的，我必须编写如下内容:

public class Hello implements RequestHandler<SNSEvent, Void>{
    public Void handleRequest(SNSEvent snsEvent, Context context) {
        ...
        return null;
    }
}

这很烦人。

是否有 RequestHandler 的替代方案对于void处理程序？:

public class Hello implements EventHandler<SNSEvent>{
    public void handleEvent(SNSEvent snsEvent, Context context) {
        ...
    }
}

Answer 1

您无需为 Lambda 入口点实现接口(interface)。您的处理程序类可以只是一个具有满足要求的签名的 POJO explained in the documentation .

例如:

package example;

import com.amazonaws.services.lambda.runtime.Context;
import com.amazonaws.services.lambda.runtime.events.SNSEvent;

public class Hello {
  public void handleEvent(SNSEvent event, Context context) {
    // Process the event
  }
}

在这种情况下，您应该使用 example.Hello::handleEvent 作为处理程序配置。

另请参阅this example from the official docs :

package example;

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

public class Hello {
    public String myHandler(int myCount, Context context) {
        LambdaLogger logger = context.getLogger();
        logger.log("received : " + myCount);
        return String.valueOf(myCount);
    }
}