我有以下JSON文本。我如何解析它以获得pageName, pagePic, post_id等的值?

{
  "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"
    }
  ]
}

当前回答

除了其他答案,我推荐这个在线开源服务jsonschema2pojo.org,它可以从json或json模式快速生成Java类,用于GSON, Jackson 1。或者Jackson 2.x。例如,如果你有:

{
   "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
         }
    ]
}

GSON的jsonschema2pojo.org生成:

@Generated("org.jsonschema2pojo")
public class Container {
    @SerializedName("pageInfo")
    @Expose
    public PageInfo pageInfo;
    @SerializedName("posts")
    @Expose
    public List<Post> posts = new ArrayList<Post>();
}

@Generated("org.jsonschema2pojo")
public class PageInfo {
    @SerializedName("pageName")
    @Expose
    public String pageName;
    @SerializedName("pagePic")
    @Expose
    public String pagePic;
}

@Generated("org.jsonschema2pojo")
public class Post {
    @SerializedName("post_id")
    @Expose
    public String postId;
    @SerializedName("actor_id")
    @Expose
    public long actorId;
    @SerializedName("picOfPersonWhoPosted")
    @Expose
    public String picOfPersonWhoPosted;
    @SerializedName("nameOfPersonWhoPosted")
    @Expose
    public String nameOfPersonWhoPosted;
    @SerializedName("message")
    @Expose
    public String message;
    @SerializedName("likesCount")
    @Expose
    public long likesCount;
    @SerializedName("comments")
    @Expose
    public List<Object> comments = new ArrayList<Object>();
    @SerializedName("timeOfPost")
    @Expose
    public long timeOfPost;
}

其他回答

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!

如果你有一些Java类(比如Message)表示JSON字符串(jsonString),你可以使用Jackson JSON库:

Message message= new ObjectMapper().readValue(jsonString, Message.class);

你可以从message对象中获取它的任何属性。

{
   "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
}

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();
    }
}

主要有两种选择……

Object mapping. When you deserialize JSON data to a number of instances of: 1.1. Some predefined classes, like Maps. In this case, you don't have to design your own POJO classes. Some libraries: org.json.simple https://www.mkyong.com/java/json-simple-example-read-and-write-json/ 1.2. Your own POJO classes. You have to design your own POJO classes to present JSON data, but this may be helpful if you are going to use them into your business logic as well. Some libraries: Gson, Jackson (see http://tutorials.jenkov.com/java-json/index.html)

映射的主要缺点是它会导致大量内存分配(以及GC压力)和CPU占用。

面向流的解析。例如,Gson和Jackson都支持这种轻量级解析。另外,您还可以查看一个自定义的、快速的、无gc的解析器示例https://github.com/anatolygudkov/green-jelly。在需要解析大量数据和对延迟敏感的应用程序中,更倾向于使用这种方式。