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

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。

其他回答

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。

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。

本页的热门答案使用了太简单的例子,比如只有一个属性的对象(例如{name: value})。我认为这个简单但真实的例子可以帮助到一些人。

这是谷歌Translate API返回的JSON:

{
  "data": 
     {
        "translations": 
          [
            {
              "translatedText": "Arbeit"
             }
          ]
     }
}

我想检索“translatedText”属性的值。“Arbeit”使用谷歌的Gson。

两种可能的方法:

Retrieve just one needed attribute String json = callToTranslateApi("work", "de"); JsonObject jsonObject = new JsonParser().parse(json).getAsJsonObject(); return jsonObject.get("data").getAsJsonObject() .get("translations").getAsJsonArray() .get(0).getAsJsonObject() .get("translatedText").getAsString(); Create Java object from JSON class ApiResponse { Data data; class Data { Translation[] translations; class Translation { String translatedText; } } } ... Gson g = new Gson(); String json =callToTranslateApi("work", "de"); ApiResponse response = g.fromJson(json, ApiResponse.class); return response.data.translations[0].translatedText;

阅读下面的博文,Java中的JSON。

这篇文章有点老了,但我仍然想回答你的问题。

步骤1:创建数据的POJO类。

步骤2:现在使用JSON创建一个对象。

Employee employee = null;
ObjectMapper mapper = new ObjectMapper();
try {
    employee =  mapper.readValue(newFile("/home/sumit/employee.json"), Employee.class);
} 
catch(JsonGenerationException e) {
    e.printStackTrace();
}

如需进一步参考,请参阅以下链接。

{
   "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
}