我正在解析XML文件中的数据并将其存储在数据库中。现在我想通过XML文件更新数据,我有一个更新数据的URL,但我没有办法在该URL上发送数据..
请帮助。
答案 0 :(得分:1)
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(4);
nameValuePairs.add(new BasicNameValuePair("latitude", "00.11"));
nameValuePairs.add(new BasicNameValuePair("longitude", "00.11"));
String url = "http://10.15.66.101:8080/LocationServer/GetLocation";
HttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost(
url);
httpPost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
// Execute HTTP Post Request
HttpResponse response = httpClient.execute(httpPost);
BusinessManager.getHandler().getLoggerUtilityObj().printMsg(
"Posting data to server");
// This is done to shutdown the previously open http connection
httpClient.getConnectionManager().shutdown();