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

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));
  }
}

其他回答

JSON是这样的:

{
   "pageInfo": {
         "pageName": "abc",
         "pagePic": "http://example.com/content.jpg"
    }
}

Java类

class PageInfo {

    private String pageName;
    private String pagePic;

    // Getters and setters
}

将此JSON转换为Java类的代码。

    PageInfo pageInfo = JsonPath.parse(jsonString).read("$.pageInfo", PageInfo.class);

Maven

<dependency>
    <groupId>com.jayway.jsonpath</groupId>
    <artifactId>json-path</artifactId>
    <version>2.2.0</version>
</dependency>

The below example shows how to read the text in the question, represented as the "jsonText" variable. This solution uses the Java EE7 javax.json API (which is mentioned in some of the other answers). The reason I've added it as a separate answer is that the following code shows how to actually access some of the values shown in the question. An implementation of the javax.json API would be required to make this code run. The full package for each of the classes required was included as I didn't want to declare "import" statements.

javax.json.JsonReader jr = 
    javax.json.Json.createReader(new StringReader(jsonText));
javax.json.JsonObject jo = jr.readObject();

//Read the page info.
javax.json.JsonObject pageInfo = jo.getJsonObject("pageInfo");
System.out.println(pageInfo.getString("pageName"));

//Read the posts.
javax.json.JsonArray posts = jo.getJsonArray("posts");
//Read the first post.
javax.json.JsonObject post = posts.getJsonObject(0);
//Read the post_id field.
String postId = post.getString("post_id");

现在,在大家对这个答案投反对票之前因为它没有使用GSON, org。json, Jackson或任何其他可用的第三方框架,它是每个问题解析所提供文本的“所需代码”的示例。我很清楚JDK 9没有考虑遵守当前标准JSR 353,因此JSR 353规范应该与任何其他第三方JSON处理实现一样对待。

这让我惊讶于它是多么简单。你可以在默认的组织中传递一个包含JSON的String给JSONObject的构造函数。json包。

JSONArray rootOfPage =  new JSONArray(JSONString);

完成了。滴麦克风。 这也适用于JSONObjects。在此之后,您可以使用对象上的get()方法查看对象的层次结构。

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

Quick-json解析器非常简单,灵活,快速,可定制。试一试

特点:

Compliant with JSON specification (RFC4627) High-Performance JSON parser Supports Flexible/Configurable parsing approach Configurable validation of key/value pairs of any JSON Hierarchy Easy to use # Very small footprint Raises developer friendly and easy to trace exceptions Pluggable Custom Validation support - Keys/Values can be validated by configuring custom validators as and when encountered Validating and Non-Validating parser support Support for two types of configuration (JSON/XML) for using quick-JSON validating parser Requires JDK 1.5 No dependency on external libraries Support for JSON Generation through object serialisation Support for collection type selection during parsing process

它可以这样使用:

JsonParserFactory factory=JsonParserFactory.getInstance();
JSONParser parser=factory.newJsonParser();
Map jsonMap=parser.parseJson(jsonString);