Question

import net.sf.json.*;

public class JSONDemo {

/**
 * @param args
 */
public static void main(String[] args) {
    JSONObject mainObj = new JSONObject();

    JSONObject jObj1 = new JSONObject();
    JSONObject jObj2 = new JSONObject();

    JSONArray jA1 = new JSONArray();        
    JSONArray jA2 = new JSONArray();

    JSONArray mainArray= new JSONArray();

    jObj1.accumulate("id", 17);
    jObj1.accumulate("name", "Alex");
    jObj1.accumulate("children", jA1);

    mainArray.add(jObj1);

    jObj2.accumulate("id", 94);
    jObj2.accumulate("name", "Steve");
    jObj2.accumulate("children", jA2);

    //Adding the new object to jObj1 via jA1

    jA1.add(jObj2);

    mainObj.accumulate("ccgs", mainArray);
    System.out.println(mainObj.toString());     
}    

}

我得到的输出是

{"ccgs":[{"id":17,"name":"Alex","children":[]}]}

我希望jObj2的子键中有jObj1。

Answer 1

显然，节点创建顺序会对生成的字符串产生影响。如果更改对象创建顺序，从子项开始，Json是正确的。

查看该代码：

    public static void main(String[] args) {
        // first create the child node
        JSONObject jObj2 = new JSONObject();
        jObj2.accumulate("id", 94);
        jObj2.accumulate("name", "Steve");
        jObj2.accumulate("children", new JSONArray());

        // then create the parent's children array
        JSONArray jA1 = new JSONArray(); 
        jA1.add(jObj2);

        // then create the parent
        JSONObject jObj1 = new JSONObject();
        jObj1.accumulate("id", 17);
        jObj1.accumulate("name", "Alex");
        jObj1.accumulate("children", jA1);

        // then create the main array
        JSONArray mainArray = new JSONArray();
        mainArray.add(jObj1);

        // then create the main object
        JSONObject mainObj = new JSONObject();
        mainObj.accumulate("ccgs", mainArray);

        System.out.println(mainObj);    
    }

输出结果为：

{"ccgs":[{"id":17,"name":"Alex","children":[{"id":94,"name":"Steve","children":[]}]}]}

Answer 2

如果你想要这样的东西{“ccgs”：[{“id”：17，“name”：“Alex”，“children”：{“id”：94，“name”：“Steve”，“儿“：[]}}]}

然后你可以这样做。

    JSONObject mainObj = new JSONObject();

    JSONObject jObj1 = new JSONObject();
    JSONObject jObj2 = new JSONObject();

    JSONArray jA1 = new JSONArray();        
    JSONArray jA2 = new JSONArray();

    JSONArray mainArray= new JSONArray();

    jObj2.accumulate("id", 94);
    jObj2.accumulate("name", "Steve");
    jObj2.accumulate("children", jA2);

    jObj1.accumulate("id", 17);
    jObj1.accumulate("name", "Alex");
    jObj1.accumulate("children", jObj2);

    mainArray.add(jObj1);



    //Adding the new object to jObj1 via jA1

    jA1.add(jObj2);

    mainObj.accumulate("ccgs", mainArray);
    System.out.println(mainObj.toString());

JSON嵌套数据创建在Java中失败

2 个答案: