如何从 DynamoDB 流新映像获取纯 Json 字符串

Question

我有一个启用了流式处理的 Dynamodb 表。此外，我还为此表创建了一个触发器，用于调用 AWS Lambda 函数。在这个lambda函数中，我正在尝试从Dynamodb流中读取新图像（修改后的Dynamodb项目），并尝试从中获取纯json字符串。我的问题是，如何获取通过流发送的 DynamoDB 项目的纯 json 字符串？我正在使用下面给出的代码片段来获取新图像，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

下面是

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（）;

---

如果你想进一步将json转换为对象，你可以使用Newtonsoft

使用Newtonsoft.Json;

TModel model = JsonConvert.DeserializeObject（jsonResult）;

Answer 3

找到了一种干净利落地的方法。从 aws-java-sdk-dynamodb-1.11.15 使用 InternalUtils.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 );

Answer 4

只是总结一下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

对于那些坚持使用地图<字符串？>的对象是纯地图，但不是属性值的人，您可以执行以下操作：

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

然后将此 DynamoDB 映射转换为纯映射（相当于最初推送到 DynamoDB 中的 json）：

asMap = InternalUtils.toSimpleMapValue(dynamoDbAttributes);

Answer 6

对于面临属性值转换问题的人，请参阅以下代码：https://github.com/aws/aws-lambda-java-libs/blob/master/aws-lambda-java-events-sdk-transformer/README.md

Map<String, AttributeValue> stringAttributeValueMap = DynamodbAttributeValueTransformer.toAttributeValueMapV1(dynamodb.getNewImage());
            List stringAttributeValueMapList = new ArrayList();
            stringAttributeValueMapList.add(stringAttributeValueMap);
            List<Item> listOfItem = InternalUtils.toItemList(stringAttributeValueMapList);
            Gson gson = new GsonBuilder().setPrettyPrinting().create();
            String updatedJSON = gson.toJson(listOfItem.get(0).asMap());

Answer 7

以下是将 DynamoDB JSON 转换为普通 JSON 的方法

/**
 * Converts DynamoDB JSON to normal JSON.
 *
 * @param map Input map of String to AttributeValue.
 * @return Returns an ObjectNode containing the normal JSON.
 */
public JsonObject toJsonObject(final Map<String, AttributeValue> map) {
        final JsonNode result = mapToJsonObject(map);
        final ObjectNode objectNode = (ObjectNode) result;
        final ObjectMapper objectMapper = new ObjectMapper();
        String recordObjectString;
        try {
            recordObjectString = objectMapper.writeValueAsString(objectNode);
        } catch (JsonProcessingException e) {
            throw new RuntimeException(e);
        }
        final JsonParser jsonParser = new JsonParser();
        final JsonObject jsonObject = jsonParser.parse(recordObjectString)
                                                .getAsJsonObject();
        return jsonObject;
    }

所以在你的情况下，像这样简单地调用下面的方法

// here record is of type DynamodbStreamRecord
toJsonObject(record.getDynamodb().getNewImage());

Answer 8

这个库可以完成这项工作：dynamoDb-marshaler

var unmarshalJson = require('dynamodb-marshaler').unmarshalJson;
console.log('jsonItem Record: %j', unmarshalJson(record.dynamodb.NewImage));