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

{
   "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
}

其他回答

你可以用谷歌Gson。

使用这个库,您只需要创建一个具有相同JSON结构的模型。然后自动填充模型。你必须调用你的变量作为你的JSON键,或者使用@SerializedName如果你想使用不同的名字。

JSON

从你的例子中:

{
    "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"
        }
    ]
}

模型

class MyModel {

    private PageInfo pageInfo;
    private ArrayList<Post> posts = new ArrayList<>();
}

class PageInfo {

    private String pageName;
    private String pagePic;
}

class Post {

    private String post_id;

    @SerializedName("actor_id") // <- example SerializedName
    private String actorId;

    private String picOfPersonWhoPosted;
    private String nameOfPersonWhoPosted;
    private String message;
    private String likesCount;
    private ArrayList<String> comments;
    private String timeOfPost;
}

解析

现在你可以使用Gson库进行解析:

MyModel model = gson.fromJson(jsonString, MyModel.class);

Gradle进口

记得在应用的Gradle文件中导入这个库

implementation 'com.google.code.gson:gson:2.8.6' // or earlier versions

自动生成模型

您可以使用这样的在线工具从JSON自动生成模型。

可以使用Apache @Model注释创建表示JSON文件结构的Java模型类,并使用它们访问JSON树中的各种元素。与其他解决方案不同,该解决方案完全没有反射,因此适用于不可能反射或开销很大的环境。

有一个示例Maven项目展示了这种用法。首先它定义了结构:

@Model(className="RepositoryInfo", properties = {
    @Property(name = "id", type = int.class),
    @Property(name = "name", type = String.class),
    @Property(name = "owner", type = Owner.class),
    @Property(name = "private", type = boolean.class),
})
final class RepositoryCntrl {
    @Model(className = "Owner", properties = {
        @Property(name = "login", type = String.class)
    })
    static final class OwnerCntrl {
    }
}

然后它使用生成的RepositoryInfo和Owner类来解析所提供的输入流,并在此过程中获取某些信息:

List<RepositoryInfo> repositories = new ArrayList<>();
try (InputStream is = initializeStream(args)) {
    Models.parse(CONTEXT, RepositoryInfo.class, is, repositories);
}

System.err.println("there is " + repositories.size() + " repositories");
repositories.stream().filter((repo) -> repo != null).forEach((repo) -> {
    System.err.println("repository " + repo.getName() + 
        " is owned by " + repo.getOwner().getLogin()
    );
})

就是这样!除此之外,这里还有一个生动的要点,展示了类似的例子以及异步网络通信。

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);

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\"}");

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。