java - 使用post方法将数据从Java android发送到Web服务器

我想将数据从Java Android发送到mysql php服务器。

这是我的按钮单击代码:

public void loginPost(View view){
      String username = usernameField.getText().toString();
      String password = passwordField.getText().toString();
      String result="";

      HttpClient httpclient = new DefaultHttpClient();
      HttpPost httppost = new HttpPost("http://geospy.zz.mu/default.php");
      try {
          List <NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
          nameValuePairs.add(new BasicNameValuePair("UserName", username));
          nameValuePairs.add(new BasicNameValuePair("PassWord", password));
          httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

          HttpResponse response = httpclient.execute(httppost);
          HttpEntity entity = response.getEntity();

          if (entity != null) {
              StringBuilder sb = new StringBuilder();
              String line;
              InputStream instream = entity.getContent();
              BufferedReader bf = new BufferedReader(new InputStreamReader(instream));
              while ((line = bf.readLine()) != null ) {
                  sb.append(line).append("\n");
              }
              result = sb.toString();
              Log.i("Read from server", result);
          }

      } catch (ClientProtocolException e) {
          e.printStackTrace();
      } catch (IOException e) {
          e.printStackTrace();
      }
      status.setText(username);

      //Intent intent = new Intent(LoginActivity.this, PembukaActivity.class);
      //startActivity(intent);
}

这是我在login.php中的代码:

<?php
    include("connect.php");

    //define $myusername and $mypassword
    $myusername = $_POST['UserName'];
    $mypassword = $_POST['PassWord'];

    //to protect mysql injection
    $myusername = stripslashes($myusername);
    $mypassword = stripslashes($mypassword);
    $myusername = mysql_real_escape_string($myusername);
    $mypassword = mysql_real_escape_string($mypassword);

    $mypassword = $mypassword;

    $sql = "SELECT ID_MEMBER FROM MEMBER WHERE USERNAME='".$myusername."' and PASSWORD= '".$mypassword."'";
    echo $sql;
    $result = mysql_query($sql);


    //mysql_num_row is counting table row
    $count = mysql_num_rows($result);
            echo "<script> alert('".$count."')</script>";

    if($count == 1)
    {
        session_start();
        $row = mysql_fetch_array($result);
        //$_SESSION['login'] = $myusername;
        $_SESSION['id_member'] = $row['id_member'];

            header('Location: login.php');
    }
    else
    {
        header('Location: default.php');
    }
?>

我在清单中添加此权限:

<uses-permission android:name="android.permission.INTERNET" />

但是我运行它后，该应用程序停止了。我不知道错误在哪里。

最佳答案

尝试在ASyncTask中进行网络连接，以使您的网络连接不在UIThread上完成，我认为这就是您崩溃的原因

像这样的东西

class TheTask extends AsyncTask<Void,Void,Void>
{



      protected void onPreExecute()
      {           super.onPreExecute();

      }

       protected Void doInBackground(Void ...params)
      {


       loginPost();//View view); // View view replace
                             // i think even having view as a parameter will crash
                             // doinbackground method you have to change it i think

      }

       protected void onPostExecute(Void result)
      {

                super.onPostExecute(result);

                    // Back to UIThread, i think handle status.setText(username);
                    // from here and take it out of your loginPost() method UI operations will
                    // crash doInBackground(Void ...params)


      }
}

然后像这样在你的代码中调用它

new TheTask().execute();

编辑:以及您所有的视图，什么都不会崩溃doinbackground方法在PreExecute和OnpostExecute上以UIOperations开头和结尾