[java] Android HttpPost:如何获得结果



2 Answers

    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。

Question

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。




您应该尝试使用HttpGet而不是HttpPost。 我有一个类似的问题,并解决了它。




你可以通过这种方式做到这一点

 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);     
}

}






Related