0

我在这里有这段代码，我从http://mobilesiri.com/json-parsing-in-android-using-android-studio/获取并修改它以供我自己使用。 

    public String makeWebServiceCall(String addr, int requestMethod) {
    URL url;
    String response = "";
    try {
        url = new URL(addr);
        HttpURLConnection conn = (HttpURLConnection) url.openConnection();
        conn.setReadTimeout(15001);
        conn.setConnectTimeout(15001);
        conn.setDoInput(true);
        conn.setDoOutput(true);

        if (requestMethod == Constant.GET) {
            conn.setRequestMethod("GET");
        } else if (requestMethod == Constant.DELETE) {
            conn.setRequestMethod("DELETE");
        }

        int reqResponseCode = conn.getResponseCode();

        if ((requestMethod == Constant.GET || requestMethod == Constant.DELETE) && reqResponseCode == HttpURLConnection.HTTP_OK) {
            String line;
            BufferedReader br = new BufferedReader(new InputStreamReader(conn.getInputStream()));
            while ((line = br.readLine()) != null) {
                response += line;
            }
        } else {
            response = "";
        }

    } catch (MalformedURLException murle) {
        Log.e(MalformedURLException.class.getName(), murle.getMessage());
    } catch (IOException ioe) {
        Log.e(IOException.class.getName(), ioe.getMessage());
    }

    return response;
}

把它和库一起放在一个普通的Java环境中，我可以得到这种形式的输出

{"status":"FOUND","data":[{"category_id":3,"category_name":"Experience Sharing Area"},{"category_id":4,"category_name":"Frequently Asked Questions"},{"category_id":1,"category_name":"General Pain Advice"},{"category_id":2,"category_name":"Pain Categorization Section"}]}

把它放在 Android 中解析数据，我得到 JSONException - org.json.JSONException: End of input at character 0 of

这是我实现它的地方。

protected Void doInBackground(Void... voids) {
    WebRequest webReq = new WebRequest();
    String jsonStr = webReq.makeWebServiceCall(Constant.WEB_SERVER_ADDR + Constant.GET_CATEGORY, Constant.GET);
    Log.d("URL:", Constant.WEB_SERVER_ADDR + Constant.GET_CATEGORY);
    Log.d("Response:", " > " + jsonStr);

    categoryList = parseJSON(jsonStr);

    return null;
}

这有什么问题吗？

4

1 回答 1

1

您收到JSONException: End of input at character 0。

发布您的parseJSON 类。

{"status":"FOUND","data":[{"category_id":3,"category_name":"经验分享区"},{"category_id":4,"category_name":"常见问题"}, {"category_id":1,"category_name":"一般疼痛建议"},{"category_id":2,"category_name":"疼痛分类部分"}]}

你的解析将是

try {
JSONObject reader = new JSONObject("YOUR_JSON_STRING");
JSONArray jsonArray = reader.getJSONArray("data");


for (int i = 0; i < jsonArray.length(); i++)
{
    JSONObject jsonItem = jsonArray.getJSONObject(i);

    try {
        jsonItem = jsonArray.getJSONObject(i);
    } catch (JSONException e1) {
        e1.printStackTrace();
    }


    String cate_id= jsonItem.getString("category_id");
    String category_name= jsonItem.getString("category_name");

}
// Add adapter

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

}
于 2016-12-12T09:58:44.053 回答