我已经创建了以下函数来从服务器获取json:
public class JSONParser {
static InputStream is = null;
static JSONObject jObj = null;
static String json = "";
// constructor
public JSONParser() {
}
public JSONObject getJSONFromUrl(String url) {
// Making HTTP request
try {
// defaultHttpClient
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost(url);
HttpResponse httpResponse = httpClient.execute(httpPost);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
} catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
try {
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");
}
is.close();
json = sb.toString();
} catch (Exception e) {
Log.e("Buffer Error", "Error converting result " + e.toString());
}
// try parse the string to a JSON object
try {
jObj = new JSONObject(json);
} catch (JSONException e) {
Log.e("JSON Parser", "Error parsing data " + e.toString());
}
// return JSON String
return jObj;
}
}
当用户连接不良时,只是在没有给出任何通知的情况下加载,我的问题是:当请求超时或连接错误时,如何添加如toast?
答案 0 :(得分:2)
首先设置HttpRequestTimeOut,如下所示,然后将此逻辑添加到try并捕获并在请求超时失败时将toast消息添加到catch块中它将显示toast消息。
url = new URI(s.replace(" ", "%20"));
Log.e("my webservice", "My webservice : " + url);
HttpGet httpget = new HttpGet(url);
HttpResponse httpResponse = null;
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);
// Execute HTTP Post Request
httpResponse = httpClient.execute(httpget);
答案 1 :(得分:2)
我知道,已经很久了,但是我试着帮忙,让我们尝试添加这个:
HttpParams httpParameters = new BasicHttpParams();
HttpConnectionParams.setConnectionTimeout(httpParameters, 10000);
HttpConnectionParams.setSoTimeout(httpParameters, 10000);
DefaultHttpClient httpClient = new DefaultHttpClient(httpParameters);
在您的代码中,这将是:
try {
HttpParams httpParameters = new BasicHttpParams();
HttpConnectionParams.setConnectionTimeout(httpParameters, 10000);
HttpConnectionParams.setSoTimeout(httpParameters, 10000);
DefaultHttpClient httpClient = new DefaultHttpClient(httpParameters);
HttpPost httpPost = new HttpPost(url);
HttpResponse httpResponse = httpClient.execute(httpPost);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();
} catch(ConnectTimeoutException e){
Log.e("Timeout Exception: ", e.toString());
} catch(SocketTimeoutException ste){
Log.e("Timeout Exception: ", ste.toString());
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
} catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
并显示消息请求超时,如果结果为null或为空,则可以在onPostExecute()
上添加警报消息。