Question

我有一个Json字符串。

每次打开我的应用程序并使用它时，我想获得一个条目。搜索网络，我创建了这样的东西：

ArrayList<HashMap<String, String>> Listads = new ArrayList<HashMap<String, String>>();
        HashMap<String, String> mapads = new HashMap<String, String>();
        String randomValue = null;
        try {
            for (adsTrend tr : objs.getTrends())

            {

                Log.i("ADS",
                        tr.getId() + " - " + tr.getLink() + " - "
                                + tr.getType() + " - " + tr.getEnabled());

                lv_arr[i] = tr.getId() + " - " + tr.getLink() + " - "
                        + tr.getType() + " - " + tr.getEnabled();
                i++;

                mapads.put("id", tr.getId());
                mapads.put("link", tr.getLink());
                mapads.put("type", tr.getType());
                mapads.put("enabled", tr.getEnabled());

                Listads.add(mapads);
                Random generator = new Random();
                Object[] values = mapads.values().toArray();
                randomValue = (String) values[generator.nextInt(values.length)];

            }



            Toast.makeText(SplashActivity.this,"this is my random value : "+randomValue,Toast.LENGTH_LONG).show();

Answer 1

假设您选择的所有内部JSON对象随机包含相同的字符串（即启用和键入），则选择随机对象很容易。您有一个嵌套JSONArray，其中包含一些内部JSONObject。

（1）提出初始回复的JSONObject

JSONObject response = new JSONObject(serverResponse);

（2）提取趋势数组

JSONArray trends = response.getJSONArray("trends");

（3）获取趋势数组的大小

int trendsSize = trends.length();

（4）选择0和数组大小之间的随机索引 - 1（包括0）

Random r = new Random();
int randomObjectIndex = r.nextInt(trendsSize-0) + 0;

应该选择一个以trendSize为界的数字（该数字不会被包括在内，因此它有效地为trendSize-1）和0

（5）获取该位置的对象

JSONObject selectedRandomObject = trends.getJSONObject(randomObjectIndex);

（6）提取你想要的字符串

String type = selectedRandomObject.getString("type");

只要您要查找的字符串存在，就不应该获得JSONException

Answer 2

试试这个

JSONObject jsonObj = new JSONObject(jsonStr); 

// using JSONArray to grab the trendsfrom under popop 
 JSONArray menuitemArr = popupObject.getJSONArray("trends");  

// lets loop through the JSONArray and get all the items 
for (int i = 0; i < menuitemArr.length(); i++) { 
   // printing the values to the logcat 
      Log.v(menuitemArr.getJSONObject(i).getString("_id").toString()); 
      Log.v(menuitemArr.getJSONObject(i).getString("_link").toString()); 
      Log.v(menuitemArr.getJSONObject(i).getString("Enabled").toString()); 
}

Answer 3

将this与您的代码结合使用：

ArrayList<HashMap<String, String>> Listads = new ArrayList<HashMap<String, String>>();
        HashMap<String, String> mapads = new HashMap<String, String>();
        String randomValue = null;
        try {
            for (adsTrend tr : objs.getTrends())

            {

                Log.i("ADS",
                        tr.getId() + " - " + tr.getLink() + " - "
                                + tr.getType() + " - " + tr.getEnabled());

                lv_arr[i] = tr.getId() + " - " + tr.getLink() + " - "
                        + tr.getType() + " - " + tr.getEnabled();
                i++;

                mapads.put("id", tr.getId());
                mapads.put("link", tr.getLink());
                mapads.put("type", tr.getType());
                mapads.put("enabled", tr.getEnabled());

                Listads.add(mapads);
                Collections.shuffle(Listads);
                Listadds.get(0);

                Log.i("ADS",
                        tr.getId() + " - " + tr.getLink() + " - "
                                + tr.getType() + " - " + tr.getEnabled());

从Json String中获取随机物品

3 个答案: