Question

我一直在寻找解决方案来解决这个问题，但我似乎无法找到解决方案。我获得的JSON数据格式与我发现的其他示例不同。

这是我从php（在Web服务器中）获得的JSON数据的示例：

{"cols":[

{"label":"time","type":"datetime"},
{"label":"Temperature","type":"number"}],

"rows":[

{"c":[{"v":"Date(2014, 9, 14, 14, 8, 50)"},{"v":23.02}]},
{"c":[{"v":"Date(2014, 9, 14, 14, 12, 1)"},{"v":23.02}]},
{"c":[{"v":"Date(2014, 9, 14, 14, 12, 3)"},{"v":23.02}]},
{"c":[{"v":"Date(2014, 9, 14, 14, 12, 5)"},{"v":23.02}]},
{"c":[{"v":"Date(2014, 9, 14, 14, 12, 7)"},{"v":22.54}]},
...

我不是创建该PHP文件的人，我无法从中更改任何内容。

我尝试使用JSONParser，就像所有示例一样，但它不起作用。如何解析数据并将其显示在ListView上？

Answer 1

您可以尝试类似以下内容：

public String getDataString() {
    String someString = "";

    JSONObject jsonObj = getData();
    Log.i("JSON string =>", jsonObj.toString());

    String text= null;

    try {
            // Handle cols
            JSONArray cols = jsonObj.getJSONArray("cols");
            int i = 0;
            Log.i("i", i+ "," + cols.length() ); //TODO delete this
            while(i<cols.length()) {

                JSONObject obj= cols.getJSONObject(i);
                String labelType = obj.getString("label");

                if(labelType.equals("time"))
                {
                    text = obj.getString("datetime");
                }
                else if(labelType.equals("Temperature"))
                {
                    text = obj.getString("number");
                }

                // Do something with String text

                i++;
            }
            // Handle rows
            JSONArray rows= jsonObj.getJSONArray("rows");
            int i = 0;
            Log.i("i", i+ "," + rows.length() ); //TODO delete this
            while(i<rows.length()) {

                JSONObject obj= rows.getJSONObject(i);
                JSONArray cObj= obj.getJSONArray("c");
                JSONObject vFirst = cObj.getJSONObject(0);
                JSONObject vSecond = cObj.getJSONObject(1);

                String firstString = vFirst.getString("v");
                String secondString = vSecond .getString("v");

                // Do something with String firstString and/or String secondString

                i++;
            }

        }

    } catch (JSONException e) {
        Log.e("testing","Failed to load JSON");
        e.printStackTrace();
    }
. 
.
.
// Return some string
return someString;
. 
.
.
}


public static JSONObject getData() {

    HttpGet httpGet = new HttpGet("YOUR_URI_HERE");
    HttpClient client = new DefaultHttpClient();
    HttpResponse response;
    StringBuilder stringBuilder = new StringBuilder();

    try {
        response = client.execute(httpGet);
        HttpEntity entity = response.getEntity();
        InputStream stream = entity.getContent();
        int b;
        while ((b = stream.read()) != -1) {
            stringBuilder.append((char) b);
        }
    } catch (ClientProtocolException e) {
    } catch (IOException e) {
    }

    JSONObject jsonObject = new JSONObject();
    try {
        jsonObject = new JSONObject(stringBuilder.toString());
    } catch (JSONException e) {
        e.printStackTrace();
    }

    return jsonObject;
}

Answer 2

您可以使用谷歌的Gson库将JSON对象直接解析为Java对象。

显示webserver将JSON数据获取到ListView中

2 个答案: