您需要使用JsonNode和来自jackson库的ObjectMapper类来获取Json树的节点。在pom.xml中添加以下依赖项以获得对Jackson类的访问权。

<!-- https://mvnrepository.com/artifact/com.fasterxml.jackson.core/jackson-databind -->
<dependency>
    <groupId>com.fasterxml.jackson.core</groupId>
    <artifactId>jackson-databind</artifactId>
    <version>2.9.5</version>
</dependency>

你应该尝试下面的代码，这将工作:

import com.fasterxml.jackson.core.JsonGenerationException;
import com.fasterxml.jackson.databind.JsonMappingException;
import com.fasterxml.jackson.databind.JsonNode;
import com.fasterxml.jackson.databind.ObjectMapper;

class JsonNodeExtractor{

    public void convertToJson(){

        String filepath = "c:\\data.json";
        ObjectMapper mapper = new ObjectMapper();
        JsonNode node =  mapper.readTree(filepath);

        // create a JsonNode for every root or subroot element in the Json String
        JsonNode pageInfoRoot = node.path("pageInfo");

        // Fetching elements under 'pageInfo'
        String pageName =  pageInfoRoot.path("pageName").asText();
        String pagePic = pageInfoRoot.path("pagePic").asText();

        // Now fetching elements under posts
        JsonNode  postsNode = node.path("posts");
        String post_id = postsNode .path("post_id").asText();
        String nameOfPersonWhoPosted = postsNode 
        .path("nameOfPersonWhoPosted").asText();
    }
}

请像这样做:

JSONParser jsonParser = new JSONParser();
JSONObject obj = (JSONObject) jsonParser.parse(contentString);
String product = (String) jsonObject.get("productId");

首先，您需要选择一个实现库来执行此操作。

用于JSON处理的Java API (JSR 353)提供了使用对象模型和流API来解析、生成、转换和查询JSON的可移植API。

参考实现在这里:https://jsonp.java.net/

下面是JSR 353的实现列表:

哪些API实现了JSR-353 (JSON)

为了帮助你决定…我也找到了这篇文章:

http://blog.takipi.com/the-ultimate-json-library-json-simple-vs-gson-vs-jackson-vs-json/

如果您选择Jackson，这里有一篇关于使用Jackson在JSON和Java之间转换的好文章:https://www.mkyong.com/java/how-to-convert-java-object-to-from-json-jackson/

希望能有所帮助!

If one wants to create Java object from JSON and vice versa, use GSON or JACKSON third party jars etc. //from object to JSON Gson gson = new Gson(); gson.toJson(yourObject); // from JSON to object yourObject o = gson.fromJson(JSONString,yourObject.class); But if one just want to parse a JSON string and get some values, (OR create a JSON string from scratch to send over wire) just use JaveEE jar which contains JsonReader, JsonArray, JsonObject etc. You may want to download the implementation of that spec like javax.json. With these two jars I am able to parse the json and use the values. These APIs actually follow the DOM/SAX parsing model of XML. Response response = request.get(); // REST call JsonReader jsonReader = Json.createReader(new StringReader(response.readEntity(String.class))); JsonArray jsonArray = jsonReader.readArray(); ListIterator l = jsonArray.listIterator(); while ( l.hasNext() ) { JsonObject j = (JsonObject)l.next(); JsonObject ciAttr = j.getJsonObject("ciAttributes");

我们可以使用JSONObject类将JSON字符串转换为JSON对象， 和遍历JSON对象。使用下面的代码。

JSONObject jObj = new JSONObject(contents.trim());
Iterator<?> keys = jObj.keys();

while( keys.hasNext() ) {
  String key = (String)keys.next();
  if ( jObj.get(key) instanceof JSONObject ) {           
    System.out.println(jObj.getString(String key));
  }
}

{
   "pageInfo": {
         "pageName": "abc",
         "pagePic": "http://example.com/content.jpg"
    },
    "posts": [
         {
              "post_id": "123456789012_123456789012",
              "actor_id": "1234567890",
              "picOfPersonWhoPosted": "http://example.com/photo.jpg",
              "nameOfPersonWhoPosted": "Jane Doe",
              "message": "Sounds cool. Can't wait to see it!",
              "likesCount": "2",
              "comments": [],
              "timeOfPost": "1234567890"
         }
    ]
}

Java code :

JSONObject obj = new JSONObject(responsejsonobj);
String pageName = obj.getJSONObject("pageInfo").getString("pageName");

JSONArray arr = obj.getJSONArray("posts");
for (int i = 0; i < arr.length(); i++)
{
    String post_id = arr.getJSONObject(i).getString("post_id");
    ......etc
}