Android HttpPost。将数据发送到PHP服务器上的表单

Question

我使用最新的Java Eclipse软件，当我运行此HttpPost代码时，模拟器崩溃。我在我的笔记本电脑上安装了uniserver，因此我将它用作服务器。

此代码应该从以前的类调用编辑文本数据，并且使用HttpPost请求这个数据上传到在线表单上各自的领域。

编辑的文本数据是3个字段：“从”，“To”和“消息”。而且我在服务器上创建的表单也有这些相同的字段来输入数据。 （“http://19x.xx.xx.xxx/androidp2p/testform.php”）其中，19x.xx.xx.xxx是我的（本地主机）IP地址。

我正确地从以前的类拉动的数据和我的代码类似于HttpPost的例子，我在网上找到，但我不知道为什么会崩溃。

我附HttpPost方法，我想看看我是否能得到任何援助。提前感谢你。

方法1：

String myBreadu, myBreadr, myBreadm; 

@Override 
protected void onCreate(Bundle savedInstanceState) { 
    // TODO Auto-generated method stub 
    super.onCreate(savedInstanceState); 
    Bundle myBasket = getIntent().getExtras(); 
    myBreadu = myBasket.getString("keyfrom"); 
    myBreadr = myBasket.getString("keyto"); 
    myBreadm = myBasket.getString("keymsg"); 
    // Create a new HttpClient and Post Header 
    HttpClient client = new DefaultHttpClient(); 
    String postURL = ("http://19x.xx.xx.xxx/androidp2p/testform.php"); 
    HttpPost post = new HttpPost(postURL); 
    try { 
     // Add the data 
     List<NameValuePair> pairs = new ArrayList<NameValuePair>(3); 
     pairs.add(new BasicNameValuePair("keysendu", myBreadu)); 
     pairs.add(new BasicNameValuePair("keysendr", myBreadr)); 
     pairs.add(new BasicNameValuePair("keysendm", myBreadm)); 
     UrlEncodedFormEntity uefe = new UrlEncodedFormEntity(pairs); 
     post.setEntity(uefe); 
     // Execute the HTTP Post Request 
     HttpResponse response = client.execute(post); 
     // Convert the response into a String 
     HttpEntity resEntity = response.getEntity(); 
     if (resEntity != null) { 
      Log.i("RESPONSE", EntityUtils.toString(resEntity)); 
     } 
    } catch (UnsupportedEncodingException uee) { 
     uee.printStackTrace(); 
    } catch (ClientProtocolException cpe) { 
     cpe.printStackTrace(); 
    } catch (IOException ioe) { 
     ioe.printStackTrace(); 
    } 

} 

方法2：

String myBreadu, myBreadr, myBreadm; 

@Override 
protected void onCreate(Bundle savedInstanceState) { 
    // TODO Auto-generated method stub 
    super.onCreate(savedInstanceState); 
    Bundle myBasket = getIntent().getExtras(); 
    myBreadu = myBasket.getString("keyfrom"); 
    myBreadr = myBasket.getString("keyto"); 
    myBreadm = myBasket.getString("keymsg"); 
    String result = null; 
    // Create a new HttpClient and Post Header 
    HttpClient client = new DefaultHttpClient(); 
    String postURL = ("http://186.45.107.129/androidp2p/testform.php"); 
    HttpPost post = new HttpPost(postURL); 
    try { 
     // Add the data 
     List<NameValuePair> pairs = new ArrayList<NameValuePair>(3); 
     pairs.add(new BasicNameValuePair("keysendu", myBreadu)); 
     pairs.add(new BasicNameValuePair("keysendr", myBreadr)); 
     pairs.add(new BasicNameValuePair("keysendm", myBreadm)); 
     UrlEncodedFormEntity uefe = new UrlEncodedFormEntity(pairs); 
     post.setEntity(uefe); 
     // Execute the HTTP Post Request 
     HttpResponse response = client.execute(post); 
     // Convert the response into a String 
     HttpEntity resEntity = response.getEntity(); 
     BufferedReader rd = new BufferedReader(new InputStreamReader(response.getEntity().getContent())); 
     String l = ""; 
     StringBuilder sb = new StringBuilder(); 
     while ((l = rd.readLine()) != null) { 
      sb.append(l + "\n"); 
     } 
     rd.close(); 
     String result = sb.toString(); // this line gives an error "Duplicate local variable result" 
    } catch (UnsupportedEncodingException uee) { 
     uee.printStackTrace(); 
    } catch (ClientProtocolException cpe) { 
     cpe.printStackTrace(); 
    } catch (IOException ioe) { 
     ioe.printStackTrace(); 
    } 

} 

这是testform.PHP

测试表单

来源：
为：
消息：

我可以补充一点好吗？我不知道我是否应该被直接发送数据到窗体或这个其他PHP页面我有..

通过我在日志中得到错误，当我尝试HttpPost的方式是：

FATAL EXCEPTION: main 

03-07 11:36:23.226: E/AndroidRuntime(1490): java.lang.RuntimeException: Unable to start activity ComponentInfo{com.project.keegan/com.project.keegan.SendPostMethod}: android.os.NetworkOnMainThreadException 

03-07 11:36:23.226: E/AndroidRuntime(1490):  at android.app.ActivityThread.performLaunchActivity(ActivityThread.java:2059) 

03-07 11:36:23.226: E/AndroidRuntime(1490):  at android.app.ActivityThread.handleLaunchActivity(ActivityThread.java:2084) 

03-07 11:36:23.226: E/AndroidRuntime(1490):  at android.app.ActivityThread.access$600(ActivityThread.java:130) 

对不起，如果这是太多的信息家伙。谢谢。

Answer 1

您需要使用AsyncTask才能完成所有网络操作。

您的网络操作会占用大量的时间，如果它是在主UI线程上完成的UI会得到响应。如果您的用户界面长时间冻结，该应用可能会被操作系统杀死。

因此的Android 4+使得强制使用后台线程执行网络操作。

把代码做的网络活动中doInBacground()和使用​​所有的AsyncTask。

这是你的AsyncTask会是什么样子：

private class SendData extends AsyncTask<String, Integer, Void> { 
     protected void doInBackground() { 
// Create a new HttpClient and Post Header 
HttpClient client = new DefaultHttpClient(); 
String postURL = ("http://19x.xx.xx.xxx/androidp2p/testform.php"); 
HttpPost post = new HttpPost(postURL); 
try { 
    // Add the data 
    List<NameValuePair> pairs = new ArrayList<NameValuePair>(3); 
    pairs.add(new BasicNameValuePair("keysendu", myBreadu)); 
    pairs.add(new BasicNameValuePair("keysendr", myBreadr)); 
    pairs.add(new BasicNameValuePair("keysendm", myBreadm)); 
    UrlEncodedFormEntity uefe = new UrlEncodedFormEntity(pairs); 
    post.setEntity(uefe); 
    // Execute the HTTP Post Request 
    HttpResponse response = client.execute(post); 
    // Convert the response into a String 
    HttpEntity resEntity = response.getEntity(); 
    if (resEntity != null) { 
     Log.i("RESPONSE", EntityUtils.toString(resEntity)); 
    } 
} catch (UnsupportedEncodingException uee) { 
    uee.printStackTrace(); 
} catch (ClientProtocolException cpe) { 
    cpe.printStackTrace(); 
} catch (IOException ioe) { 
    ioe.printStackTrace(); 
} 
        
 } 

 protected void onProgressUpdate() { 
    //called when the background task makes any progress 
 } 

  protected void onPreExecute() { 
     //called before doInBackground() is started 
 } 
 protected void onPostExecute() { 
     //called after doInBackground() has finished  
 } 
  } 

，并且可以使用new SendData().execute("");