Question

我正在尝试解析来自this url的JSON数据。

但是我收到了这些错误：

03-27 16：48：21.019：E / Buffer Error（23717）：转换结果错误java.lang.NullPointerException

03-27 16：48：21.059：E / JSON Parser（23717）：解析数据时出错org.json.JSONException：
字符0的输入结束

当我调试我的代码时; getJsonFromUrl()方法返回 null jobject 。这是我使用的 JSONParser类。导致错误的原因是什么？

public class JSONParser {

    static InputStream iStream = null;
    static JSONArray jarray = null;
    static JSONObject jObj= null;
    static String json = "";

    public JSONParser() {
    }



    public JSONObject getJSONFromUrl(String url) {

        // Making HTTP request
        DefaultHttpClient httpClient = new DefaultHttpClient();
        HttpPost httpPost = new HttpPost(url);



        try {
            HttpResponse httpResponse = httpClient.execute(httpPost);
            HttpEntity httpEntity = httpResponse.getEntity();
            InputStream is = httpEntity.getContent();
            BufferedReader reader = new BufferedReader(new InputStreamReader(is, "iso-8859-1"), 8);
            StringBuilder sb = new StringBuilder();
            String line = null;
            while ((line = reader.readLine()) != null) {
                sb.append(line + "\n");
            }
            iStream.close();
            json = sb.toString();
        } catch (Exception e) {
            Log.e("Buffer Error", "Error converting result " + e.toString());
        }

        // try parsing the string to a JSON object
        try {
            if (json != null) {
                jObj = new JSONObject(json);
            } else {
                jObj = null;
            }
        } catch (JSONException e) {
            Log.e("JSON Parser", "Error parsing data " + e.toString());
        }

        // return JSON String
        return jObj;

    }
}

我正在使用这些行从另一个类调用此方法。（url参数在顶部定义）

  JSONParser jParser = new JSONParser();
  final JSONObject jobject = jParser.getJSONFromUrl(url);

Answer 1

您正在尝试使用HTTP POST方法而不是相应的GET方法（W3schools.com GET vs.POST）来获取JSON内容，修改源代码以简化和修复您的HTTP请求

DefaultHttpClient httpClient = new DefaultHttpClient();
HttpGet get = new HttpGet(url);

try {
    HttpResponse httpResponse = httpClient.execute(get);
    String json = EntityUtils.toString(httpResponse.getEntity());
    System.out.println(json);
    ....
    ....

} catch (Exception e) {
    ....
}

解析数据时出错org.json.JSONException：字符0处的输入结束

1 个答案: