Android ListView:循环遍历json数组列表

时间:2013-07-22 04:21:54

标签: android

现在已经工作了几个小时了,不太确定我应该如何对我的测试阵列进行排序。我已经尝试了一些我见过的不同选项,这个是关于最简单的实现到我的项目中。任何可能帮助我克服这个驼峰的建议或想法?

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
}

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;
    }
}

1 个答案:

答案 0 :(得分:1)

您可以执行以下操作:在DecimalFormat循环之外定义新的for

DecimalFormat myFormatter = new DecimalFormat("0000"); 

然后在for循环内使用格式化程序形成id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);

只有当您的ids最多为4位时,上述操作才有效。您可以使用"Count-end"值而不是使用硬编码"0000"来使格式动态化。

如果"Questions"对象是JSONArray,则可以避免所有这些麻烦。