Question

请记住JSON结构事先不知道，即它完全是任意的，我们只知道它是JSON格式。

例如，

以下JSON

{
   "Port":
   {
       "@alias": "defaultHttp",
       "Enabled": "true",
       "Number": "10092",
       "Protocol": "http",
       "KeepAliveTimeout": "20000",
       "ThreadPool":
       {
           "@enabled": "false",
           "Max": "150",
           "ThreadPriority": "5"
       },
       "ExtendedProperties":
       {
           "Property":
           [                         
               {
                   "@name": "connectionTimeout",
                   "$": "20000"
               }
           ]
       }
   }
}

应该反序列化为具有键的类似Map的结构（为了简洁，不包括以上所有内容）：

port[0].alias
port[0].enabled
port[0].extendedProperties.connectionTimeout
port[0].threadPool.max

我目前正在调查杰克逊，所以我们有：

TypeReference<HashMap<String, Object>> typeRef = new TypeReference<HashMap<String, Object>>() {};
Map<String, String> o = objectMapper.readValue(jsonString, typeRef);

但是，生成的Map实例基本上是嵌套地图的地图：

{Port={@alias=diagnostics, Enabled=false, Type=DIAGNOSTIC, Number=10033, Protocol=JDWP, ExtendedProperties={Property={@name=suspend, $=n}}}}

虽然我需要使用“点符号”来展平平面地图，如上所述。

我宁愿不自己实现这个，虽然目前我没有看到任何其他方式......

Answer 1

你可以这样做来遍历树，并跟踪你有多深，找出点符号属性名称：

import com.fasterxml.jackson.databind.JsonNode;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.node.ArrayNode;
import com.fasterxml.jackson.databind.node.ObjectNode;
import com.fasterxml.jackson.databind.node.ValueNode;
import java.io.IOException;
import java.util.HashMap;
import java.util.Iterator;
import java.util.Map;
import org.junit.Test;

public class FlattenJson {
  String json = "{\n" +
      "   \"Port\":\n" +
      "   {\n" +
      "       \"@alias\": \"defaultHttp\",\n" +
      "       \"Enabled\": \"true\",\n" +
      "       \"Number\": \"10092\",\n" +
      "       \"Protocol\": \"http\",\n" +
      "       \"KeepAliveTimeout\": \"20000\",\n" +
      "       \"ThreadPool\":\n" +
      "       {\n" +
      "           \"@enabled\": \"false\",\n" +
      "           \"Max\": \"150\",\n" +
      "           \"ThreadPriority\": \"5\"\n" +
      "       },\n" +
      "       \"ExtendedProperties\":\n" +
      "       {\n" +
      "           \"Property\":\n" +
      "           [                         \n" +
      "               {\n" +
      "                   \"@name\": \"connectionTimeout\",\n" +
      "                   \"$\": \"20000\"\n" +
      "               }\n" +
      "           ]\n" +
      "       }\n" +
      "   }\n" +
      "}";

  @Test
  public void testCreatingKeyValues() {
    Map<String, String> map = new HashMap<String, String>();
    try {
      addKeys("", new ObjectMapper().readTree(json), map);
    } catch (IOException e) {
      e.printStackTrace();
    }
    System.out.println(map);
  }

  private void addKeys(String currentPath, JsonNode jsonNode, Map<String, String> map) {
    if (jsonNode.isObject()) {
      ObjectNode objectNode = (ObjectNode) jsonNode;
      Iterator<Map.Entry<String, JsonNode>> iter = objectNode.fields();
      String pathPrefix = currentPath.isEmpty() ? "" : currentPath + ".";

      while (iter.hasNext()) {
        Map.Entry<String, JsonNode> entry = iter.next();
        addKeys(pathPrefix + entry.getKey(), entry.getValue(), map);
      }
    } else if (jsonNode.isArray()) {
      ArrayNode arrayNode = (ArrayNode) jsonNode;
      for (int i = 0; i < arrayNode.size(); i++) {
        addKeys(currentPath + "[" + i + "]", arrayNode.get(i), map);
      }
    } else if (jsonNode.isValueNode()) {
      ValueNode valueNode = (ValueNode) jsonNode;
      map.put(currentPath, valueNode.asText());
    }
  }
}

它产生以下地图：

Port.ThreadPool.Max=150, 
Port.ThreadPool.@enabled=false, 
Port.Number=10092, 
Port.ExtendedProperties.Property[0].@name=connectionTimeout, 
Port.ThreadPool.ThreadPriority=5, 
Port.Protocol=http, 
Port.KeepAliveTimeout=20000, 
Port.ExtendedProperties.Property[0].$=20000, 
Port.@alias=defaultHttp, 
Port.Enabled=true

在属性名称中删除@和$应该很容易，但是因为你说JSON是任意的，所以你最终可能会在键名中发生冲突。

Answer 2

如何使用json-flattener。 https://github.com/wnameless/json-flattener

BTW，我是这个lib的作者。

String flattenedJson = JsonFlattener.flatten(yourJson);
Map<String, Object> flattenedJsonMap = JsonFlattener.flattenAsMap(yourJson);

// Result:
{
    "Port.@alias":"defaultHttp",
    "Port.Enabled":"true",
    "Port.Number":"10092",
    "Port.Protocol":"http",
    "Port.KeepAliveTimeout":"20000",
    "Port.ThreadPool.@enabled":"false",
    "Port.ThreadPool.Max":"150",
    "Port.ThreadPool.ThreadPriority":"5",
    "Port.ExtendedProperties.Property[0].@name":"connectionTimeout",
    "Port.ExtendedProperties.Property[0].$":"20000"
}

Answer 3

怎么样：

import java.util.ArrayList;
import java.util.HashMap;
import java.util.Map;
import java.util.Map.Entry;
import com.google.gson.Gson;

/**
 * NOT FOR CONCURENT USE
*/
@SuppressWarnings("unchecked")
public class JsonParser{

Gson gson=new Gson();
Map<String, String> flatmap = new HashMap<String, String>();

public Map<String, String> parse(String value) {        
    iterableCrawl("", null, (gson.fromJson(value, flatmap.getClass())).entrySet());     
    return flatmap; 
}

private <T> void iterableCrawl(String prefix, String suffix, Iterable<T> iterable) {
    int key = 0;
    for (T t : iterable) {
        if (suffix!=null)
            crawl(t, prefix+(key++)+suffix);
        else
            crawl(((Entry<String, Object>) t).getValue(), prefix+((Entry<String, Object>) t).getKey());
    }
}

private void crawl(Object object, String key) {
    if (object instanceof ArrayList)
        iterableCrawl(key+"[", "]", (ArrayList<Object>)object);
    else if (object instanceof Map)
        iterableCrawl(key+".", null, ((Map<String, Object>)object).entrySet());
    else
        flatmap.put(key, object.toString());
}
}

Answer 4

您可以使用Typesafe Config Library实现类似的功能，如下例所示：

import com.typesafe.config.*;
import java.util.Map;
public class TypesafeConfigExample {
  public static void main(String[] args) {
    Config cfg = ConfigFactory.parseString(
      "   \"Port\":\n" +
      "   {\n" +
      "       \"@alias\": \"defaultHttp\",\n" +
      "       \"Enabled\": \"true\",\n" +
      "       \"Number\": \"10092\",\n" +
      "       \"Protocol\": \"http\",\n" +
      "       \"KeepAliveTimeout\": \"20000\",\n" +
      "       \"ThreadPool\":\n" +
      "       {\n" +
      "           \"@enabled\": \"false\",\n" +
      "           \"Max\": \"150\",\n" +
      "           \"ThreadPriority\": \"5\"\n" +
      "       },\n" +
      "       \"ExtendedProperties\":\n" +
      "       {\n" +
      "           \"Property\":\n" +
      "           [                         \n" +
      "               {\n" +
      "                   \"@name\": \"connectionTimeout\",\n" +
      "                   \"$\": \"20000\"\n" +
      "               }\n" +
      "           ]\n" +
      "       }\n" +
      "   }\n" +
      "}");

    // each key has a similar form to what you need
    for (Map.Entry<String, ConfigValue> e : cfg.entrySet()) {
      System.out.println(e);
    }
  }
}

Answer 5

来自org.springframework.integration.transformer.ObjectToMapTransformer的

Spring Integration会产生预期的结果。默认情况下，它将shouldFlattenKeys属性设置为true并生成平面贴图（不嵌套，值始终为简单类型）。当shouldFlattenKeys=false生成嵌套地图时

ObjectToMapTransformer旨在用作集成流程的一部分，但以独立方式使用它是完全正确的。您需要使用转换输入的有效负载构造org.springframework.messaging.Message。 transform方法返回org.springframework.messaging.Message对象，其有效负载为Map

import org.springframework.integration.transformer.ObjectToMapTransformer;
import org.springframework.messaging.Message;
import org.springframework.messaging.support.GenericMessage;

Message message = new GenericMessage(value);
 ObjectToMapTransformer transformer = new ObjectToMapTransformer();
        transformer.setShouldFlattenKeys(true);
        Map<String,Object> payload = (Map<String, Object>) transformer
                .transform(message)
                .getPayload();

旁注：将Spring Integration添加到类路径只是为了使用单个类可能有点过头了，但是你可以检查这个类的实现并自己编写类似的解决方案。嵌套地图由杰克逊（org.springframework.integration.support.json.JsonObjectMapper#fromJson(payload, Map.class)）生成，然后mapis以递归方式进行，将所有值作为集合展平。

Answer 6

如果您事先知道结构，则可以定义Java类并使用gson将JSON解析为该类的实例：

YourClass obj = gson.fromJson(json, YourClass.class);

如果没有，那么我不确定你要做什么。您显然无法动态定义类，因此使用点符号访问已解析的JSON是不可能的。

除非你想要这样的东西：

Map<String, String> parsed = magicParse(json);
parsed["Port.ThreadPool.max"]; // returns 150

如果是这样，那么遍历你的地图并制作一个“扁平”的地图似乎不是太大的问题。

还是别的什么？

Answer 7

我也不得不在我的项目中解决一个类似的问题，并发现 springframework.vault 有一个方法 flatten() 来做同样的事情。下面是一个示例代码。


    //Json string to Map<String, Object>

    String data = "Your json as string"
    final ObjectMapper mapper = new ObjectMapper();
    final MapType type = mapper.getTypeFactory().constructMapType(
                Map.class, String.class, Object.class);
    final Map<String, Object> map = mapper.readValue(data, type);

    //Using springframework.vault flatten method

    Map<String, String> keyMap = JsonMapFlattener.flattenToStringMap(map);

    //Input

    {"key": {"nested": 1}, "another.key": ["one", "two"] }

    //Output

      key.nested=1
      another.key[0]=one
      another.key[1]=two

记得添加依赖

    <dependency>
        <groupId>org.springframework.vault</groupId>
        <artifactId>spring-vault-core</artifactId>
        <version>2.1.1.RELEASE</version>
    </dependency>

有关详细信息，请参阅 https://docs.spring.io/spring-vault/docs/current/api/org/springframework/vault/support/JsonMapFlattener.html

Answer 8

import com.alibaba.fastjson.JSONObject;
import com.alibaba.fastjson.JSONPath;

import java.util.Iterator;
import java.util.Map;

public class JsonMapFlattener {


    public static Map<String, Object> flatten(Map<String, ? extends Object> inputMap) {
        return org.springframework.vault.support.JsonMapFlattener.flatten(inputMap);
    }


    public static JSONObject unflatten(Map<String, ? extends Object> inputMap) {
        Map<String, Object> map = flatten(inputMap);
        Iterator<String> it = map.keySet().iterator();
        JSONObject jo2 = new JSONObject();
        while (it.hasNext()) {
            String key = it.next();
            Object value = map.get(key);
            JSONPath.set(jo2, key, value);
        }
        return jo2;

    }
}

如何将JSON反序列化为平面，类似Map的结构？

8 个答案: