Question

我启用了一个支持流媒体的Dynamodb表。此外，我还为此表创建了一个触发AWS Lambda函数的触发器。在这个lambda函数中，我尝试从Dynamodb流中读取新图像（修改后的Dynamodb项）并尝试从中获取纯json字符串。我的问题是如何才能获得通过流发送的DynamoDB项的纯json字符串？我使用下面给出的代码片段来获取新的图像，但我不知道如何从中获取json字符串。感谢您的帮助。

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.";
}

}

Answer 1

以下是从Dynamo JSON转换为标准JSON的完整代码：

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.";
    }
}

Answer 2

在C＃中，您可以使用DynamoDB Document类将newImage转换为纯json

使用Amazon.DynamoDBv2.DocumentModel;

var streamRecord = dynamoEvent.Records.First（）;

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

如果您想进一步将json转换为对象，则可以使用Newtonsoft

使用Newtonsoft.Json;

TModel模型= JsonConvert.DeserializeObject（jsonResult）;

Answer 3

找到一种干净利落的方式。使用aws-java-sdk-dynamodb-1.11.15.jar中的InternalUtils

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 );

Answer 4

总结Himanshu Parmar的答案：

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();
}

Answer 5

你有没有办法做到这一点。一种粗略的方法是创建自己的解析器，但即使我们不想采用这种方法

Answer 6

对于那些坚持使用 Map 对象是普通 Map 而不是 Attributes 值的人，您可以执行以下操作：

Map<String, AttributeValue> dynamoDbAttributes = 
    objectMapper.convertValue(dynamoDbMap, new TypeReference<Map<String, AttributeValue>>() {});

然后把这个DynamoDB Map转换成一个普通的Map（相当于最初推入DynamoDb的json）：

asMap = InternalUtils.toSimpleMapValue(dynamoDbAttributes);

Answer 7

这个图书馆完成工作：dynamoDb-marshaler

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

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

如何从DynamoDB流新的图像中获取纯Json字符串？

7 个答案: