Question

我有一个json对象，如下所示

{
"Book1":{"Title1":"Cost1","Title2":"Cost2"}, 
"Book2":{"Title3 ":"Cost3"},
"Book3":{"Title4 ":"Cost4","Title5 ":"Cost5"},
"Book4":{"Title6":"Cost6","Title7”:”Cost7","Title8”:”Cost8"}
}

我想要检索＆＃34;标题＆＃34;的所有值没有任何关键字，我想在一个数组中有{Title1，Title2，Title3，Title4，...}。

下面是我根据Book名称检索的代码，但现在我需要检索所有值

    List<String> books=new ArrayList<String >();
    try {
        JSONObject obj = new JSONObject(loadJSONFromAssetArea());
        JSONObject bookJson = obj.getJSONObject("Book1");


        Iterator<String> keys= bookJson.keys();
        do
        {
            String keyValue = (String)keys.next();
            JSONArray specTitle = obj.getJSONArray(keyValue);
            for (int i = 0; i < specTitle.length(); i++) {
               books.add(specTitle.get(i).toString());
            }

        }while(keys.hasNext());


    } catch (JSONException e) {
        e.printStackTrace();
    }

我没有得到如何将键值传递给json对象。请帮助。

提前致谢

Answer 1

注意

您的 GIVEN Json 无效。

Book1,Book2.. DYNAMIC/UNKNOWN 。

以这种方式使用 Iterator 。

试试这种方式，

JSONObject jOBJ= new JSONObject(loadJSONFromAssetArea()); Iterator iteratorObj = jOBJ.keys(); while (iteratorObj.hasNext()) { String getJsonObj = (String)iteratorObj.next(); System.out.println("Key: " + Key + "------>" + getJsonObj); // print Book1,Book2.. // Now your work }

Answer 2

如果你想解析这个

{
"Book1":{"Title1":"Cost1","Title2":"Cost2"}, 
"Book2":{"Title3 ":"Cost3"},
"Book3":{"Title4 ":"Cost4","Title5 ":"Cost5"},
"Book4":{"Title6":"Cost6","Title7”:”Cost7","Title8”:”Cost8"}
}

您可以使用此代码来解析json对象，

List<String> books = new ArrayList<String>();
    try {
        JSONObject obj = new JSONObject(loadJSONFromAssetArea());
        JSONObject bookJson, innerObject;
        Iterator<String> keys = obj.keys();
        do {
            String keyValue = (String) keys.next();
            if (obj.get(keyValue) instanceof JSONObject) {
                bookJson = obj.getJSONObject(keyValue);
                Iterator<?> innerKeys = bookJson.keys();
                while (innerKeys.hasNext()) {
                    String innerKey = (String) innerKeys.next();
                    innerObject = bookJson.getJSONObject(innerKey);
                    books.add(innerObject.getString(innerKey));
                }
            }
        } while (keys.hasNext());
    } catch (JSONException e) {
        e.printStackTrace();
    }

Answer 3

尝试这是有效的

 JSONObject jsonObject = new JSONObject(loadJSONFromAssetArea());

 List<String> books=new ArrayList<String >();
        try {
               Iterator  iteratorObj = jsonObject.keys();
            while (iteratorObj.hasNext())
            {
                String getJsonObj = (String)iteratorObj.next();
                System.out.println("Book No.: "   + "------>" + getJsonObj);

                JSONObject jo_inside = jsonObject.getJSONObject(getJsonObj);
                //JSONObject jo_inside = new JSONObject((String) iteratorObj.next());

                Iterator<String> keys = jo_inside.keys();
                while (keys.hasNext())
                {
                    String key = keys.next();
                    String value = jo_inside.getString(key);
                    Log.v("Book Title key", key);
                    Log.v("Book Name value", value);
                    books.add(jo_inside.getString(key));
                }

            }

    } catch (JSONException e) {
            e.printStackTrace();
        }

将json对象的所有值检索到数组列表中

3 个答案: