我遇到了一个简单的问题
在我的活动中,我进行了一个异步函数调用
json = jsonParser.post(URL, params);
我的class +函数看起来像这样(部分基于this answer by kuester2000)
public class json {
static InputStream stream = null;
static JSONObject jObj = null;
static String json = "";
public json() {
// TODO Auto-generated constructor stub
}
public JSONObject post(String url, List<NameValuePair> NameValuePairs) {
try {
String paramString = URLEncodedUtils
.format(NameValuePairs, "utf-8");
Log.d("AsyncTask","1:Startet http");
HttpParams httpParameters = new BasicHttpParams();
// Set the timeout in milliseconds until a connection is established.
// The default value is zero, that means the timeout is not used.
int timeoutConnection = 3000;
HttpConnectionParams.setConnectionTimeout(httpParameters, timeoutConnection);
// Set the default socket timeout (SO_TIMEOUT)
// in milliseconds which is the timeout for waiting for data.
int timeoutSocket = 5000;
HttpConnectionParams.setSoTimeout(httpParameters, timeoutSocket);
DefaultHttpClient httpClient = new DefaultHttpClient(httpParameters);
HttpPost httpPost = new HttpPost(url);
Log.d("AsyncTask","1: http Post");
httpPost.setEntity(new UrlEncodedFormEntity(NameValuePairs));
Log.d("AsyncTask","1: send http.post");
HttpResponse httpResponse = httpClient.execute(httpPost);
Log.d("AsyncTask", " receive Response");
// make sure response is good (Int 200)
if(httpResponse.getStatusLine().getStatusCode() == 200){
HttpEntity httpEntity = httpResponse.getEntity();
if (httpEntity != null) {
Log.d("AsyncTask", " get Entity");
stream = httpEntity.getContent();
json = convertStreamToString(stream);
Log.d("AsyncTask", "Convert Json");
stream.close();
} else {
Log.d("AsyncTask","1: no internet connection");
}
Log.d("AsyncTask", "After Convert Json");
} else {
Log.e("AsyncTask: ", "server not found error code: " + String.valueOf(httpResponse.getStatusLine().getStatusCode()));
}
} catch (UnsupportedEncodingException e) {
Log.d("AsyncTask","2: no internet connection");
e.printStackTrace();
} catch (ClientProtocolException e) {
Log.d("AsyncTask","3: no internet connection");
e.printStackTrace();
} catch (IOException e) {
Log.d("AsyncTask","7: no internet connection");
e.printStackTrace();
}
try {
jObj = new JSONObject(json);
} catch (JSONException e) {
Log.e("AsyncTask","error1");
//Log.e("JSON Parser", "Error parsing data " + e.toString());
}
Log.e("AsyncTask","retun to getgame");
//Log.d("JSON_POST", jObj.toString());
// return JSON String
return jObj; // <--- this is the problem
}
我的问题是,当服务器没有响应ok时,我的函数返回null。(使应用程序终止)我想以某种方式捕捉到这一点,并返回到我的异步,在那里我想发布一个带有错误代码的吐司......以便用户知道稍后再试。
有人能帮我怎么做吗?
1条答案
按热度按时间wnrlj8wa1#
与盲目返回
jObj
不同,您可以尝试检查它是否为null,如果是,则为其分配一个新的空白JSONObject
。将最后一行替换为:这样至少可以消除null错误。