首先，您需要选择一个实现库来执行此操作。

用于JSON处理的Java API (JSR 353)提供了使用对象模型和流API来解析、生成、转换和查询JSON的可移植API。

参考实现在这里:https://jsonp.java.net/

下面是JSR 353的实现列表:

哪些API实现了JSR-353 (JSON)

为了帮助你决定…我也找到了这篇文章:

http://blog.takipi.com/the-ultimate-json-library-json-simple-vs-gson-vs-jackson-vs-json/

如果您选择Jackson，这里有一篇关于使用Jackson在JSON和Java之间转换的好文章:https://www.mkyong.com/java/how-to-convert-java-object-to-from-json-jackson/

希望能有所帮助!

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。

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));
  }
}

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}]}

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

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。