通過Android應用程序連接到PHP腳本

Question

這是我到目前爲止的代碼。

public void postData(String toPost) { 
    // Create a new HttpClient and Post Header 
    HttpClient httpclient = new DefaultHttpClient(); 
    HttpPost httppost = new HttpPost("http://www.mywebsite.com/dev/reverser.php"); 

    //This is the data to send 
    String MyName = toPost; //any data to send 

    try { 
    // Add your data 
    List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(1); 
    nameValuePairs.add(new BasicNameValuePair("action", MyName)); 

    httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs)); 

    // Execute HTTP Post Request 

    ResponseHandler<String> responseHandler = new BasicResponseHandler(); 
    String response = httpclient.execute(httppost, responseHandler); 

    //This is the response from a php application 

    String reverseString = response; 
    Toast.makeText(this, "response" + reverseString, Toast.LENGTH_LONG).show(); 

    } catch (ClientProtocolException e) { 
    Toast.makeText(this, "CPE response " + e.toString(), Toast.LENGTH_LONG).show(); 
    // TODO Auto-generated catch block 
    } catch (IOException e) { 
    Toast.makeText(this, "IOE response " + e.toString(), Toast.LENGTH_LONG).show(); 
    // TODO Auto-generated catch block 
    } 

    }//end postData() 

有人能告訴我下面的代碼有什麼問題！我已經確定只有try catch塊存在問題，而不是活動中的其他任何地方。我只是不知道它是什麼或如何糾正它。

我的PHP代碼很簡單。這是這樣的 -

//code to reverse the string 
$reversed = strrev($_POST["action"]); 
echo $reversed; 

Answer 1

添加爲海報要求這樣做的單獨答案。

假設：

一個）有一個允許的URL的文本框加載

b）中點擊它時，A按鈕上的URL中取出˚F

實現一個按鈕執行聯網操作點擊偵聽器，調用以下函數：

private void URL() 
    { 
     String url = txtURL.getText().toString(); 
     new URLTask().execute(new String[] { url }); 
    }  




private class URLTask extends AsyncTask<String, Void, String> 
    { 
     protected String doInBackground(String... urls) 
     { 

      BufferedReader br = null; 
      String url = urls[0]; 
      StringBuffer sb = new StringBuffer(""); 

      try 
      { 
       HttpClient client = new DefaultHttpClient(); 
       HttpPost request = new HttpPost(); 
       request.setURI(new URI(url)); 

       List<NameValuePair> postParameters = new ArrayList<NameValuePair>(); 
       postParameters.add(new BasicNameValuePair("param1", "value of param1")); 

       UrlEncodedFormEntity formEntity = new UrlEncodedFormEntity(postParameters);   

       request.setEntity(formEntity); 
       HttpResponse response = client.execute(request); 

       String line; 

       br = new BufferedReader(new InputStreamReader(response.getEntity().getContent())); 
       while ((line = br.readLine()) != null) 
       { 
        sb.append(line + "\n"); 
       } 
       br.close(); 

      } 
      catch(Exception e) 
      { 
       Toast.makeText(HttpPostActivity.this, "Exception: " + e.toString(), Toast.LENGTH_LONG).show(); 
      } 
      finally 
      { 
       if (br != null) 
       { 
        try 
        { 
         br.close(); 
        } 
        catch(IOException ioe) 
        { 
         ioe.printStackTrace(); 
        } 
       }   

      }   

      return(sb.toString()); 
     } 


     protected void onPostExecute(String result) 
     { 
      txtContent.setText(result); 
     } 

您還需要實現onPostExecute。還有其他APIS。

Android Async Task Documentation

Answer 2

嘗試打印出例外。 使用此代碼打印出您的回覆。檢查回覆的狀態

 HttpResponse response = client.execute(request); 

     String line; 

     br = new BufferedReader(new InputStreamReader(response.getEntity().getContent())); 
     while ((line = br.readLine()) != null) 
     { 
      sb.append(line + "\n"); 
     } 
     br.close(); 

    } 
    catch(Exception e) 
    { 
     Toast.makeText(HttpPostActivity.this, "Exception: " + e.toString(), Toast.LENGTH_LONG).show(); 
    } 

告訴我們例外情況。那麼很容易找出你的問題。

編輯答案：

您正在嘗試在主線程中進行聯網操作。這是一件非法的事情。創建一個AsyncTask，即創建一個獨立的線程來完成您的網絡操作。

Android Details of the exception

Stackoverflow question

Answer 3

在你上面的註釋表明，你得到一個android.os.NetworkOnMainThreadException。

在Android的最新版本中，您不允許在主線程上進行聯網，因爲它會導致UI無響應。使用AsyncTask（see the Android developer guide for details）或其他一些機制將代碼移動到其他線程。