org。Json库易于使用。

只要记住(在强制转换或使用getJSONObject和getJSONArray等方法时)JSON表示法

[…]表示一个数组，因此库将把它解析为JSONArray {…}表示一个对象，因此库将把它解析为JSONObject

示例代码如下:

import org.json.*;

String jsonString = ... ; //assign your JSON String here
JSONObject obj = new JSONObject(jsonString);
String pageName = obj.getJSONObject("pageInfo").getString("pageName");

JSONArray arr = obj.getJSONArray("posts"); // notice that `"posts": [...]`
for (int i = 0; i < arr.length(); i++)
{
    String post_id = arr.getJSONObject(i).getString("post_id");
    ......
}

你可以从以下几个方面找到更多的例子

可下载的jar: http://mvnrepository.com/artifact/org.json/json

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。

您可以使用JsonNode来表示JSON字符串的结构化树。它是无处不在的杰克逊图书馆的一部分。

ObjectMapper mapper = new ObjectMapper();
JsonNode yourObj = mapper.readTree("{\"k\":\"v\"}");

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;
}

为了便于示例，让我们假设您有一个只有名称的Person类。

private class Person {
    public String name;

    public Person(String name) {
        this.name = name;
    }
}

谷歌GSON (Maven)

我个人最喜欢的JSON对象序列化/反序列化。

Gson g = new Gson();

Person person = g.fromJson("{\"name\": \"John\"}", Person.class);
System.out.println(person.name); //John

System.out.println(g.toJson(person)); // {"name":"John"}

更新

如果你想获取单个属性，你可以很容易地使用谷歌库:

JsonObject jsonObject = new JsonParser().parse("{\"name\": \"John\"}").getAsJsonObject();

System.out.println(jsonObject.get("name").getAsString()); //John

Org。JSON (Maven)

如果您不需要对象反序列化，而只是获得一个属性，您可以尝试org。json(或查看上面的GSON示例!)

JSONObject obj = new JSONObject("{\"name\": \"John\"}");

System.out.println(obj.getString("name")); //John

杰克逊(Maven)

ObjectMapper mapper = new ObjectMapper();
Person user = mapper.readValue("{\"name\": \"John\"}", Person.class);

System.out.println(user.name); //John

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));
  }
}