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

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\"}");

其他回答

Jsoniter (jsoniterator)是一个相对较新的和简单的json库,旨在简单和快速。反序列化json数据所需要做的就是

JsonIterator.deserialize(jsonData, int[].class);

其中jsonData是json数据的字符串。

去官方网站看看吧 获取更多信息。

{
   "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
}

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。

如果你的数据很简单,你不想要外部依赖,可以使用以下几行代码:

/**
 * A very simple JSON parser for one level, everything quoted.
 * @param json the json content.
 * @return a key => value map.
 */
public static Map<String, String> simpleParseJson(String json) {
    Map<String, String> map = new TreeMap<>();
    String qs[] = json.replace("\\\"", "\u0001").replace("\\\\", "\\").split("\"");
    for (int i = 1; i + 3 < qs.length; i += 4) {
        map.put(qs[i].replace('\u0001', '"'), qs[i + 2].replace('\u0001', '"'));
    }
    return map;
}

这些数据

{"name":"John", "age":"30", "car":"a \"quoted\" back\\slash car"}

生成一个包含

{age=30, car=a "quoted" back\slash car, name=John}

这也可以升级为使用未加引号的值…

/**
 * A very simple JSON parser for one level, names are quoted.
 * @param json the json content.
 * @return a key => value map.
 */
public static Map<String, String> simpleParseJson(String json) {
    Map<String, String> map = new TreeMap<>();
    String qs[] = json.replace("\\\"", "\u0001").replace("\\\\",  "\\").split("\"");
    for (int i = 1; i + 1 < qs.length; i += 4) {
        if (qs[i + 1].trim().length() > 1) {
            String x = qs[i + 1].trim();
            map.put(qs[i].replace('\u0001', '"'), x.substring(1, x.length() - 1).trim().replace('\u0001', '"'));
            i -= 2;
        } else {
            map.put(qs[i].replace('\u0001', '"'), qs[i + 2].replace('\u0001', '"'));
        }
    }
    return map;
}

为了解决复杂的结构,它变得很难看… ... 对不起! !... 但我忍不住把它编码了^^ 这将解析给定的JSON以及更多内容。它产生嵌套的映射和列表。

/**
 * A very simple JSON parser, names are quoted.
 * 
 * @param json the json content.
 * @return a key => value map.
 */
public static Map<String, Object> simpleParseJson(String json) {
    Map<String, Object> map = new TreeMap<>();
    String qs[] = json.replace("\\\"", "\u0001").replace("\\\\", "\\").split("\"");
    int index[] = { 1 };
    recurse(index, map, qs);
    return map;
}

/**
 * Eierlegende Wollmilchsau.
 * 
 * @param index index into array.
 * @param map   the current map to fill.
 * @param qs    the data.
 */
private static void recurse(int[] index, Map<String, Object> map, String[] qs) {
    int i = index[0];
    for (;; i += 4) {
        String end = qs[i - 1].trim(); // check for termination of an object
        if (end.startsWith("}")) {
            qs[i - 1] = end.substring(1).trim();
            i -= 4;
            break;
        }

        String key = qs[i].replace('\u0001', '"');
        String x = qs[i + 1].trim();
        if (x.endsWith("{")) {
            x = x.substring(0, x.length() - 1).trim();
            if (x.endsWith("[")) {
                List<Object> list = new ArrayList<>();
                index[0] = i + 2;
                for (;;) {
                    Map<String, Object> inner = new TreeMap<>();
                    list.add(inner);
                    recurse(index, inner, qs);
                    map.put(key, list);
                    i = index[0];

                    String y = qs[i + 3]; // check for termination of array
                    if (y.startsWith("]")) {
                        qs[i + 3] = y.substring(1).trim();
                        break;
                    }
                }
                continue;
            }

            Map<String, Object> inner = new TreeMap<>();
            index[0] = i + 2;
            recurse(index, inner, qs);
            map.put(key, inner);
            i = index[0];
            continue;
        }
        if (x.length() > 1) { // unquoted
            String value = x.substring(1, x.length() - 1).trim().replace('\u0001', '"');
            if ("[]".equals(value)) // handle empty array
                map.put(key, new ArrayList<>());
            else
                map.put(key, value);
            i -= 2;
        } else {
            map.put(key, qs[i + 2].replace('\u0001', '"'));
        }
    }
    index[0] = i;
}

yield -如果你打印地图:

{pageInfo={pageName=abc, pagePic=http://example.com/content.jpg}, posts=[{actor_id=1234567890, comments=[], likesCount=2, message=Sounds cool. Can't wait to see it!, nameOfPersonWhoPosted=Jane Doe, picOfPersonWhoPosted=http://example.com/photo.jpg, post_id=123456789012_123456789012, timeOfPost=1234567890}]}

如果你有一些Java类(比如Message)表示JSON字符串(jsonString),你可以使用Jackson JSON库:

Message message= new ObjectMapper().readValue(jsonString, Message.class);

你可以从message对象中获取它的任何属性。