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

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

其他回答

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;
}

请像这样做:

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

由于还没有人提到它,这里是一个使用Nashorn (Java 8的JavaScript运行时部分,但在Java 11中已弃用)的解决方案的开始。

解决方案

private static final String EXTRACTOR_SCRIPT =
    "var fun = function(raw) { " +
    "var json = JSON.parse(raw); " +
    "return [json.pageInfo.pageName, json.pageInfo.pagePic, json.posts[0].post_id];};";

public void run() throws ScriptException, NoSuchMethodException {
    ScriptEngine engine = new ScriptEngineManager().getEngineByName("nashorn");
    engine.eval(EXTRACTOR_SCRIPT);
    Invocable invocable = (Invocable) engine;
    JSObject result = (JSObject) invocable.invokeFunction("fun", JSON);
    result.values().forEach(e -> System.out.println(e));
}

性能比较

我编写的JSON内容包含三个数组,分别为20、20和100个元素。我只想从第三个数组中获取100个元素。我使用下面的JavaScript函数来解析和获取我的条目。

var fun = function(raw) {JSON.parse(raw).entries};

使用Nashorn运行一百万次调用需要7.5~7.8秒

(JSObject) invocable.invokeFunction("fun", json);

org。Json需要20~21秒

new JSONObject(JSON).getJSONArray("entries");

杰克逊用时6.5~7秒

mapper.readValue(JSON, Entries.class).getEntries();

在这种情况下,Jackson的性能比Nashorn好,后者的性能比org.json好得多。 Nashorn API比org更难使用。json或Jackson的。根据您的需求,Jackson和Nashorn都是可行的解决方案。

{
   "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
}