13

I need to convert a AWS DYNAMODB JSON to a standard JSON object. so I can remove the data type from the DynamoDB JSON Something more like:

in DYNAMODB JSON:

"videos": [
    {
      "file": {
        "S": "file1.mp4"
      },
      "id": {
        "S": "1"
      },
      "canvas": {
        "S": "This is Canvas1"
      }
    },
    {
      "file": {
        "S": "main.mp4"
      },
      "id": {
        "S": "0"
      },
      "canvas": {
        "S": "this is a canvas"
      }
    }
  ]

to Standard JSON
 "videos": [
    {
      "file": "file1.mp4"
      ,
      "id": "1"
      ,
      "canvas":  "This is Canvas1"
      ,
      "file": "main.mp4"
      ,
      "id":  "0"
      ,
      "canvas": "this is a canvas"

    }
  ]

I found a nice tool in Javascript but is there any tool in Java in order to do that?

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1

7 Answers 7

Reset to default
21

You can use ItemUtils class in aws sdk. Below is sample code using Kotlin:

import com.amazonaws.services.dynamodbv2.document.ItemUtils
import com.amazonaws.services.dynamodbv2.model.AttributeValue

fun main(args: Array<String>) {
    val data = HashMap<String,AttributeValue>()
    data.put("hello",AttributeValue().withS("world"))
    println(data.toString())
    println(ItemUtils.toItem(data).toJSON())
}

Output:

{hello={S: world,}}
{"hello":"world"}
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3 Comments

Nice and helpful. The exact solution which I was looking for. Just need to keep in mind that it works only for new aws versions (not sure which version exactly but didn't work until I upgraded to the latest while before that I was using version from December 2017)
this solution don't work with Java SDK 2.X dependencies for AWS. The ItemUtils class is not implemented here implementation platform('software.amazon.awssdk:bom:2.17.100') implementation 'software.amazon.awssdk:dynamodb'
This answer is no longer valid; see stackoverflow.com/questions/71262197/…
4

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.";
    }
}
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Comments

4

Following is a simple solution which can be applied to convert any DynamoDB Json to Simple JSON.

//passing the reponse.getItems() 
public static Object getJson(List<Map<String,AttributeValue>> mapList) {
    List<Object> finalJson= new ArrayList();
    for(Map<String,AttributeValue> eachEntry : mapList) {
        finalJson.add(mapToJson(eachEntry));
    }
    return finalJson;
}


//if the map is null then it add the key and value(string) in the finalKeyValueMap
public static Map<String,Object> mapToJson(Map<String,AttributeValue> keyValueMap){
    Map<String,Object> finalKeyValueMap = new HashMap();
    for(Map.Entry<String, AttributeValue> entry : keyValueMap.entrySet()) 
    {
        if(entry.getValue().getM() == null) {
            finalKeyValueMap.put(entry.getKey(),entry.getValue().getS());
        }
        else {
            finalKeyValueMap.put(entry.getKey(),mapToJson(entry.getValue().getM()));
        }
    }
    return finalKeyValueMap;
}

This will produce your desired Json in the form of List>. Then using the Gson you can convert it into the Json format. 

Gson gson = new Gson();
String jsonString = gson.toJson(getJson(response.getItems()));
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Comments

3

Convert your JSON string to a Map first using

JSONObject jsonObj = new JSONObject(logString);
HashMap<String, Object> myMap = new Gson().fromJson(jsonObj.toString(), HashMap.class);

Function for converting DynamoDB JSON to Standard JSON

public static Map<String,Object> mapToJson(Map map){
        Map<String,Object> finalKeyValueMap = new HashMap();
        Iterator it = map.entrySet().iterator();

        while(it.hasNext()) {
            Map.Entry pair = (Map.Entry)it.next();

            Map obj1 = (Map) pair.getValue();
            if(obj1.get("m") == null) {
                if(obj1.get("n") != null)
                    finalKeyValueMap.put(pair.getKey().toString(),obj1.get("n"));
                else if(obj1.get("s") != null)
                    finalKeyValueMap.put(pair.getKey().toString(),obj1.get("s"));
            }
            else {
                Map obj2 = (Map) pair.getValue();
                Map obj3 = (Map) obj2.get("m");
                finalKeyValueMap.put(pair.getKey().toString(),mapToJson(obj3));
            }

        }
        System.out.println(finalKeyValueMap.toString());
        return finalKeyValueMap;
    }

calling mapToJson(myMap); will return a standard Map which you can convert back to JSON

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Comments

0

Amazon has an internal utility that can do the core of the job in 1 line.

Map<String, Object> jsonMapWithId = InternalUtils.toSimpleMapValue(attributeValueMap);

A long version which contains some configurations:

import com.amazonaws.services.dynamodbv2.document.internal.InternalUtils;
import com.amazonaws.services.dynamodbv2.model.AttributeValue;
import java.lang.Map;
import java.lang.HashMap;

public class JsonConvert {
    public static void main(String[] args) {
        Map<String,Object> jsonMap = new HashMap();
        jsonMap.put("string", "foo");
        jsonMap.put("number", 123);
        Map<String,Object> innerMap = new HashMap();
        innerMap.put("key", "value");
        jsonMap.put("map", innerMap);
        AttributeValue attributeValue = InternalUtils.toAttributeValue(jsonMap);
        Map<String, AttributeValue> attributeValueMap = new HashMap();
        attributeValueMap.put("id", attributeValue);

        Map<String, Object> jsonMapWithId = InternalUtils.toSimpleMapValue(attributeValueMap);
        Map<String, Object> jsonMap = jsonMapWithId.get("id"); // This is a regular map, values are not Amazon AttributeValue

        Gson gson = new Gson(); 
        String json = gson.toJson(jsonMap); 

        System.out.println(json);
    }
}
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1 Comment

This no longer works in current SDK. See stackoverflow.com/questions/71262197/…
0

There is now EnhancedDocument interface to address this https://aws.amazon.com/blogs/devops/introducing-the-enhanced-document-api-for-dynamodb-in-the-aws-sdk-for-java-2-x/

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1 Comment

Whilst this may theoretically answer the question, it would be preferable to include the essential parts of the answer here, and provide the link for reference. Please edit the answer with all relevant information. Make sure to use your own words, answers comprised entirely of a quote (sourced or not) will often be deleted since they do not contain any original content.
-1 
import com.amazonaws.services.dynamodbv2.document.ItemUtils;
import com.amazonaws.services.dynamodbv2.model.AttributeValue;
import com.fasterxml.jackson.core.type.TypeReference;
import com.fasterxml.jackson.databind.JsonNode;
import com.fasterxml.jackson.databind.MapperFeature;
import com.fasterxml.jackson.databind.ObjectMapper;

import java.io.IOException;
import java.util.Map;

public final class Utility {

  public static <T> String convertDynamoDbJsonToNormalJson(final JsonNode node, final Class<T> type)
      throws IOException {
    final var objectMapper =
        new ObjectMapper().configure(MapperFeature.ACCEPT_CASE_INSENSITIVE_PROPERTIES, true);

    final Map<String, AttributeValue> itemMap =
        objectMapper.convertValue(node, new TypeReference<>() {});
    final var item = ItemUtils.toItem(itemMap);
    final var asPojo = objectMapper.readValue(item.toJSON(), type);
    return objectMapper.writeValueAsString((Object) asPojo);
  }
}
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2 Comments

This is answer is no longer valid in current versions of the SDK, see stackoverflow.com/questions/71262197/…
Please read How to Answer and edit your answer to contain an explanation as to why this code would actually solve the problem at hand. Always remember that you're not only solving the problem, but are also educating the OP and any future readers of this post.

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