0

现在已经为此工作了几个小时，不太确定我应该如何对我的测试数组进行排序。我已经尝试了一些不同的选项，我已经看到发布这个选项是最简单的实现到我的项目中。有什么建议或想法可以帮助我克服这个困难吗？

来自 php 的 JSON 数组

{
    "Questions": {
        "0001": {
            "Title": "What is Stackoverflow",
            "Answer": "C",
            "User": "testUSER",
            "Date": "0000-00-00",
            "Used": "0"
        },
        "0002": {
            "Title": "What is Stackoverflow",
            "Answer": "C",
            "User": "testUSERb",
            "Date": "0000-00-00",
            "Used": "1"
        },
        "0003": {
            "Title": "What is Stackoverflow",
            "Answer": "C",
            "User": "testUSERc",
            "Date": "0000-00-00",
            "Used": "0"
        }
    },
    "Count-start": 1,
    "Count-end": 3
}

已使用：0 = 未使用 1 = 用户已使用。我将使用它来确定是否向用户显示问题和答案。

助手类

public JSONObject query(){
        HttpClient client = new DefaultHttpClient();
        HttpConnectionParams.setConnectionTimeout(client.getParams(), 10000);
        HttpResponse response;
        JSONObject json = new JSONObject();
        HttpPost post = new HttpPost(QUERY_LINK);
        post.setHeader("json", json.toString());
        StringEntity se;
        try {
            se = new StringEntity(json.toString());
            se.setContentEncoding(new BasicHeader(HTTP.CONTENT_TYPE,"application/json"));
            post.setEntity(se);
            response = client.execute(post);
            if (response != null) {
                InputStream in = response.getEntity().getContent();
                a = convertStreamToString(in);
                JSONObject check = new JSONObject(a);   
                return check;
            }
        } catch (UnsupportedEncodingException e) {
            e.printStackTrace();
        } catch (ClientProtocolException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        } catch (JSONException e) {
            e.printStackTrace();
        }

        return json;
    }

主要活动

private ArrayList<ListHelper> GetSearchResults(){
        JSONObject getJSON = new Helper().query();
        JSONArray jArrayObject = new JSONArray();
        jArrayObject.put(getJSON);
        int end = 0,start = 0;
        String title, answer, user, used;
        ListHelper sr = new ListHelper();


     ArrayList<ListHelper> results = new ArrayList<ListHelper>();
     JSONObject offerObject = null;
        try {
            offerObject = getJSON.getJSONObject("Questions");
            start = offerObject.getInt("Count-start");
            end = offerObject.getInt("Count-end");
        } catch (JSONException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }


     for(int a = start; a < end; a = a++) {
         try {


            JSONObject businessObject = offerObject.getJSONObject("Id");
            // this is were i'm stuck, how to go about selecting each id

            title = businessObject.getString("Title");
            answer = businessObject.getString("Answer");
            user = businessObject.getString("User");
            used = businessObject.getString("Used");

            sr = new ListHelper();
             sr.setTitle(title);
             sr.setUser(user);
             sr.setUsed(used);
             results.add(sr);
        } catch (JSONException e1) {
            e1.printStackTrace();
        }
      }
     return results;
    }
}
4

1 回答 1

1

您可以为您的目的执行以下操作：在循环DecimalFormat外定义一个新的for

DecimalFormat myFormatter = new DecimalFormat("0000");

然后在for循环内id使用格式化程序形成您的字符串。

String id = myFormatter.format(a);

然后使用id获取JSONObject

// this is were i'm stuck, how to go about selecting each id
JSONObject businessObject = offerObject.getJSONObject(id);

仅当您ids最多 4 位数字时，上述内容才有效。"Count-end"您可以通过使用值而不是使用硬编码来使格式动态化"0000"。

"Questions"如果对象是 JSONArray，那么所有这些麻烦都可以避免。

于 2013-07-22T05:33:13.953 回答