Android将发布数据发送到webservice
这个问题在这里已经有了答案：
使用参数调用sendPostRequest(String username, String pass)方法。 您可以从UI线程调用此方法，请求将从不同的线程（嵌入AsyncTask ）发送。
private void sendPostRequest(String givenUsername, String givenPassword) {
class SendPostReqAsyncTask extends AsyncTask<String, Void, String> {
@Override
protected String doInBackground(String... params) {
String paramUsername = params[0];
String paramPassword = params[1];
System.out.println("*** doInBackground ** paramUsername "
+ paramUsername + " paramPassword :" + paramPassword);
HttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost(
"http://lib-dm.process9.com/libertydm/ValidateUserHandler.ashx");// replace with your url
httpPost.addHeader("Content-type",
"application/x-www-form-urlencoded");
BasicNameValuePair usernameBasicNameValuePair = new BasicNameValuePair(
"UserId", paramUsername); // Make your own key value pair
BasicNameValuePair passwordBasicNameValuePAir = new BasicNameValuePair(
"Password", paramPassword);// make your own key value pair
// You can add more parameters like above
List<NameValuePair> nameValuePairList = new ArrayList<NameValuePair>();
nameValuePairList.add(usernameBasicNameValuePair);
nameValuePairList.add(passwordBasicNameValuePair);
try {
UrlEncodedFormEntity urlEncodedFormEntity = new UrlEncodedFormEntity(
nameValuePairList);
httpPost.setEntity(urlEncodedFormEntity);
try {
HttpResponse httpResponse = httpClient
.execute(httpPost);
InputStream inputStream = httpResponse.getEntity()
.getContent();
InputStreamReader inputStreamReader = new InputStreamReader(
inputStream);
BufferedReader bufferedReader = new BufferedReader(
inputStreamReader);
StringBuilder stringBuilder = new StringBuilder();
String bufferedStrChunk = null;
while ((bufferedStrChunk = bufferedReader.readLine()) != null) {
stringBuilder.append(bufferedStrChunk);
}
return stringBuilder.toString();
} catch (ClientProtocolException cpe) {
System.out
.println("First Exception coz of HttpResponese :"
+ cpe);
cpe.printStackTrace();
} catch (IOException ioe) {
System.out
.println("Second Exception coz of HttpResponse :"
+ ioe);
ioe.printStackTrace();
}
} catch (UnsupportedEncodingException uee) {
System.out
.println("An Exception given because of UrlEncodedFormEntity argument :"
+ uee);
uee.printStackTrace();
}
return null;
}
@Override
protected void onPostExecute(String result) {
super.onPostExecute(result);
}
}
SendPostReqAsyncTask sendPostReqAsyncTask = new SendPostReqAsyncTask();
sendPostReqAsyncTask.execute(givenUsername, givenPassword);
}
尝试实施这个代码，它应该工作。 如果你的Android版本高于4.0，那么你将不得不强制使用Asyn Task，这样这个http请求就不会在Main Thread上运行。 否则应用程序将崩溃。 如果应用程序使用HTTPGET或HTTPPOST方法并且正在主线程上运行，它将会崩溃。 Asyn TASk是解决方案。
HttpClient client = new DefaultHttpClient();
HttpResponse response;
JSONObject json=new JSONObject();
HttpPost post = new HttpPost(url1);
try {
json.put("Key","your value");
json.put("Key", "Value");
StringEntity stringEntity = new StringEntity(json.toString());
stringEntity.setContentEncoding("UTF-8");
stringEntity.setContentType("application/json");
post.setEntity(stringEntity);
response = client.execute(post);
Log.e("RESPONSE", response.toString());
String responseBody = EntityUtils
.toString(response.getEntity());
res= responseBody.toString();
Log.e("RESPONSE BODY", responseBody);
看看这个和这个。
您不能在主线程上执行网络操作。 一点Google搜索总是可以帮助你。
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