我一直在尝试发送HttpPost请求并检索响应，但即使我能够建立连接，我还没有得到如何获取请求 - 响应返回的字符串消息
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("http://www.myurl.com/app/page.php");
// Add your data
List < NameValuePair > nameValuePairs = new ArrayList < NameValuePair > (5);
nameValuePairs.add(new BasicNameValuePair("type", "20"));
nameValuePairs.add(new BasicNameValuePair("mob", "919895865899"));
nameValuePairs.add(new BasicNameValuePair("pack", "0"));
nameValuePairs.add(new BasicNameValuePair("exchk", "1"));
try {
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
Log.d("myapp", "works till here. 2");
try {
HttpResponse response = httpclient.execute(httppost);
Log.d("myapp", "response " + response.getEntity());
} catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
}
对不起，我听起来很天真，因为我是java的新手。请帮帮我。
答案 0 :(得分：99)
尝试在回复中使用EntityUtil：
String responseBody = EntityUtils.toString(response.getEntity());
答案 1 :(得分：8)
URL url;
url = new URL("http://www.url.com/app.php");
URLConnection connection;
connection = url.openConnection();
HttpURLConnection httppost = (HttpURLConnection) connection;
httppost.setDoInput(true);
httppost.setDoOutput(true);
httppost.setRequestMethod("POST");
httppost.setRequestProperty("User-Agent", "Tranz-Version-t1.914");
httppost.setRequestProperty("Accept_Language", "en-US");
httppost.setRequestProperty("Content-Type",
"application/x-www-form-urlencoded");
DataOutputStream dos = new DataOutputStream(httppost.getOutputStream());
dos.write(b); // bytes[] b of post data
String reply;
InputStream in = httppost.getInputStream();
StringBuffer sb = new StringBuffer();
try {
int chr;
while ((chr = in.read()) != -1) {
sb.append((char) chr);
}
reply = sb.toString();
} finally {
in.close();
}
此代码段有效。 我在搜索之后得到了它，但是来自J2ME代码。
答案 2 :(得分：7)
您可以使用ResponseHandler调用execute方法。这是一个例子：
ResponseHandler<String> responseHandler = new BasicResponseHandler();
String response = httpClient.execute(httppost, responseHandler);
答案 3 :(得分：4)
你可以这样做：
public class MyHttpPostProjectActivity extends Activity implements OnClickListener {
private EditText usernameEditText;
private EditText passwordEditText;
private Button sendPostReqButton;
private Button clearButton;
/** Called when the activity is first created. */
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.login);
usernameEditText = (EditText) findViewById(R.id.login_username_editText);
passwordEditText = (EditText) findViewById(R.id.login_password_editText);
sendPostReqButton = (Button) findViewById(R.id.login_sendPostReq_button);
sendPostReqButton.setOnClickListener(this);
clearButton = (Button) findViewById(R.id.login_clear_button);
clearButton.setOnClickListener(this);
}
@Override
public void onClick(View v) {
if(v.getId() == R.id.login_clear_button){
usernameEditText.setText("");
passwordEditText.setText("");
passwordEditText.setCursorVisible(false);
passwordEditText.setFocusable(false);
usernameEditText.setCursorVisible(true);
passwordEditText.setFocusable(true);
}else if(v.getId() == R.id.login_sendPostReq_button){
String givenUsername = usernameEditText.getEditableText().toString();
String givenPassword = passwordEditText.getEditableText().toString();
System.out.println("Given username :" + givenUsername + " Given password :" + givenPassword);
sendPostRequest(givenUsername, givenPassword);
}
}
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();
// In a POST request, we don't pass the values in the URL.
//Therefore we use only the web page URL as the parameter of the HttpPost argument
HttpPost httpPost = new HttpPost("http://www.nirmana.lk/hec/android/postLogin.php");
// Because we are not passing values over the URL, we should have a mechanism to pass the values that can be
//uniquely separate by the other end.
//To achieve that we use BasicNameValuePair
//Things we need to pass with the POST request
BasicNameValuePair usernameBasicNameValuePair = new BasicNameValuePair("paramUsername", paramUsername);
BasicNameValuePair passwordBasicNameValuePAir = new BasicNameValuePair("paramPassword", paramPassword);
// We add the content that we want to pass with the POST request to as name-value pairs
//Now we put those sending details to an ArrayList with type safe of NameValuePair
List<NameValuePair> nameValuePairList = new ArrayList<NameValuePair>();
nameValuePairList.add(usernameBasicNameValuePair);
nameValuePairList.add(passwordBasicNameValuePAir);
try {
// UrlEncodedFormEntity is an entity composed of a list of url-encoded pairs.
//This is typically useful while sending an HTTP POST request.
UrlEncodedFormEntity urlEncodedFormEntity = new UrlEncodedFormEntity(nameValuePairList);
// setEntity() hands the entity (here it is urlEncodedFormEntity) to the request.
httpPost.setEntity(urlEncodedFormEntity);
try {
// HttpResponse is an interface just like HttpPost.
//Therefore we can't initialize them
HttpResponse httpResponse = httpClient.execute(httpPost);
// According to the JAVA API, InputStream constructor do nothing.
//So we can't initialize InputStream although it is not an interface
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 caz of HttpResponese :" + cpe);
cpe.printStackTrace();
} catch (IOException ioe) {
System.out.println("Second Exception caz 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);
if(result.equals("working")){
Toast.makeText(getApplicationContext(), "HTTP POST is working...", Toast.LENGTH_LONG).show();
}else{
Toast.makeText(getApplicationContext(), "Invalid POST req...", Toast.LENGTH_LONG).show();
}
}
}
SendPostReqAsyncTask sendPostReqAsyncTask = new SendPostReqAsyncTask();
sendPostReqAsyncTask.execute(givenUsername, givenPassword);
}
}
答案 4 :(得分：1)
试试这个，它似乎是最紧凑的。虽然在现实世界中，您需要使用异步请求，以便在检索请求的页面时设备不会挂起。
http://www.softwarepassion.com/android-series-get-post-and-multipart-post-requests/