Question

我需要将AWS DYNAMODB JSON转换为标准JSON对象。所以我可以从DynamoDB JSON中删除数据类型更像是：

DYNAMODB JSON中的

：

"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"

    }
  ]

我在Javascript中找到了一个很好的工具，但是为了做到这一点，Java中是否有任何工具？

Answer 1

您可以在aws sdk中使用ItemUtils类。以下是使用Kotlin的示例代码：

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())
}

输出：

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

Answer 2

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 3

以下是一个简单的解决方案，可用于将任何DynamoDB Json转换为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;
}

这将以List＆gt;的形式产生您想要的Json。然后使用Gson，您可以将其转换为Json格式。

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

Answer 4

Amazon有一个内部实用程序，可以在1行中完成核心任务。

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

包含一些配置的长版

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);
    }
}

Answer 5

首先使用

将JSON字符串转换为Map。

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

用于将DynamoDB JSON转换为标准JSON的功能

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;
    }

调用mapToJson(myMap);将返回一个标准Map，您可以将其转换回JSON

Answer 6

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);
  }
}

使用Java将DynamoDB JSON转换为标准JSON

6 个答案: