How to get the pure Json string from DynamoDB stream new image?

问题

I've a Dynamodb table with streaming enabled. Also I've created a trigger for this table which calls an AWS Lambda function. Within this lambda function, I'm trying read the new image (Dynamodb item after the modification) from the Dynamodb stream and trying to get the pure json string out of it. My Question is how can i get the pure json string of the DynamoDB item that's been sent over the stream? I'm using the code snippet given below to get the new Image, but I've no clue how to get the json string out of it. Appreciate your help.

public class LambdaFunctionHandler implements RequestHandler<DynamodbEvent, Object> {

@Override
public Object handleRequest(DynamodbEvent input, Context context) {
    context.getLogger().log("Input: " + input);

    for (DynamodbStreamRecord record : input.getRecords()){

        context.getLogger().log(record.getEventID());
        context.getLogger().log(record.getEventName());
        context.getLogger().log(record.getDynamodb().toString());
        Map<String,AttributeValue> currentRecord = record.getDynamodb().getNewImage();

        //how to get the pure json string of the new image
        //..............................................
     }
     return "Successfully processed " + input.getRecords().size() + " records.";
}

}

回答1:

Below is the complete code for converting from Dynamo JSON to Standard JSON:

import com.amazonaws.services.dynamodbv2.document.Item;
import com.amazonaws.services.dynamodbv2.document.internal.InternalUtils;
import com.amazonaws.services.dynamodbv2.model.AttributeValue;
import com.amazonaws.services.lambda.runtime.Context;
import com.amazonaws.services.lambda.runtime.RequestHandler;
import com.amazonaws.services.lambda.runtime.events.DynamodbEvent;
import com.amazonaws.services.lambda.runtime.events.DynamodbEvent.DynamodbStreamRecord;
import com.google.gson.Gson;

import java.util.ArrayList;
import java.util.List;
import java.util.Map;

/**
 * Main Lambda class to receive event stream, parse it to Survey
 * and process them.
 */
public class SurveyEventProcessor implements
        RequestHandler<DynamodbEvent, String> {

    private static final String INSERT = "INSERT";

    private static final String MODIFY = "MODIFY";

    public String handleRequest(DynamodbEvent ddbEvent, Context context) {

        List<Item> listOfItem = new ArrayList<>();
        List<Map<String, AttributeValue>> listOfMaps = null;
        for (DynamodbStreamRecord record : ddbEvent.getRecords()) {

            if (INSERT.equals(record.getEventName()) || MODIFY.equals(record.getEventName())) {
                listOfMaps = new ArrayList<Map<String, AttributeValue>>();
                listOfMaps.add(record.getDynamodb().getNewImage());
                listOfItem = InternalUtils.toItemList(listOfMaps);
            }

            System.out.println(listOfItem);
            try {
               // String json = new ObjectMapper().writeValueAsString(listOfItem.get(0));
                Gson gson = new Gson();
                Item item = listOfItem.get(0);

                String json = gson.toJson(item.asMap());
                System.out.println("JSON is ");
                System.out.println(json);
            }catch (Exception e){
                e.printStackTrace();
            }
        }


        return "Successfully processed " + ddbEvent.getRecords().size() + " records.";
    }
} 

回答2:

In c# you can convert newImage to pure json by use of DynamoDB Document class

using Amazon.DynamoDBv2.DocumentModel;

var streamRecord = dynamoEvent.Records.First();

var jsonResult=Document.FromAttributeMap(streamRecord.Dynamodb.NewImage).ToJson();

---

and if you want to go further ahead to convert json to object you can use Newtonsoft

using Newtonsoft.Json;

TModel model = JsonConvert.DeserializeObject(jsonResult);

回答3:

Found a way of doing it cleanly. Using InternalUtils from aws-java-sdk-dynamodb-1.11.15.jar

com.amazonaws.services.dynamodbv2.model.Record streamRecord = ((RecordAdapter) record).getInternalObject();
            // get order ready //
            OrderFinal order = Utils.mapO2Object(
                    InternalUtils.toSimpleMapValue(streamRecord.getDynamodb().getNewImage().get("document").getM()), 
                    OrderFinal.class );

回答4:

Just summarizing the answer of Himanshu Parmar:

Map<String, AttributeValue> newImage = record.getDynamodb().getNewImage();
List<Map<String, AttributeValue>> listOfMaps = new ArrayList<Map<String, AttributeValue>>();
listOfMaps.add(newImage);
List<Item> itemList = ItemUtils.toItemList(listOfMaps);
for (Item item : itemList) {
    String json = item.toJSON();
}

回答5:

Did you figure out a way to do this. One crude way is to create your own parser but even we don't want to go with that approach

回答6:

This library do the job: dynamoDb-marshaler

var unmarshalJson = require('dynamodb-marshaler').unmarshalJson;

console.log('jsonItem Record: %j', unmarshalJson(record.dynamodb.NewImage));

来源：https://stackoverflow.com/questions/37655755/how-to-get-the-pure-json-string-from-dynamodb-stream-new-image