發出簡單的HTTP請求並接收內容

Question

我發現了一個工作代碼，用於在這裏製作簡單的HTTP請求，從How can I make a simple HTTP request in MainActivity.java? (Android Studio)開始，我將在下面發佈它（有一些更改，如果我沒有錯，現在需要使用try{} catch{}）。但是我想問一下如何獲得內容？我在下面的方式與代碼打交道：

GetUrlContentTask req = new GetUrlContentTask(); 
req.execute("http://192.168.1.10/?pin=OFF1"); 
textView3.setText(req.doInBackground("http://192.168.1.10/?pin=OFF1")); 

GetUrlContentTask

private class GetUrlContentTask extends AsyncTask<String, Integer, String> { 
    protected String doInBackground(String... urls) { 
     // String content1 = ""; 
     try { 
      URL url = new URL(urls[0]); 
      HttpURLConnection connection = (HttpURLConnection) url.openConnection(); 
      connection.setRequestMethod("GET"); 
      connection.setDoOutput(true); 
      connection.connect(); 


      BufferedReader rd = new BufferedReader(new InputStreamReader(connection.getInputStream())); 
      String content = "", line; 
      while ((line = rd.readLine()) != null) { 
       content += line + "\n"; 
      } 
      // content1 = content; 
     } 
     catch (Exception e) { 
      e.printStackTrace(); 
     } 

     // return content1; - returns "", wrong 
     return "aaa"; 
     //does not work return content; 
    } 

    protected void onProgressUpdate(Integer... progress) { 
    } 

    protected void onPostExecute(String result) { 
     // this is executed on the main thread after the process is over 
     // update your UI here  
    } 
} 

Answer 1

添加到您的onPostExecute(String result)通話

protected void onPostExecute(String result) { 
     // this is executed on the main thread after the process is over 
     // update your UI here  
     setMyData(result); 
    } 

然後填寫你的文本字段或任何其它數據都需要更新。

private void setMyData(String result){ 
     textView3.setText(result); 
    } 

Answer 2

你到底在哪裏卡住了？您的代碼大部分都是正確的，但您可能需要稍微重新排列它。你的範圍是有點關閉，你的評論代碼幾乎到達那裏。見下面

protected String doInBackground(String... urls) { 
    String content = "", line = ""; 
    HttpURLConnection httpURLConnection; 
    BufferedReader bufferedReader; 
    URL url; 

    try { 
     for(String uri : urls) { 
      url = new URL(uri); 
      url = new URI(url.toURI().getScheme(), url.getAuthority(), url.getPath(), "pin=" + URLEncoder.encode("&OFF1", "UTF-8"), null).toURL(); 

      httpURLConnection = (HttpURLConnection) url.openConnection(); 
      httpURLConnection.setRequestMethod("GET"); 
      httpURLConnection.setDoOutput(true); 
      httpURLConnection.connect(); 


      bufferedReader = new BufferedReader(new InputStreamReader(httpURLConnection.getInputStream())); 
      while ((line = bufferedReader.readLine()) != null) { 
       content += line + System.lineSeparator(); 
      } 
     } 
    } catch (Exception e) { 
     e.printStackTrace(); 
    } finally { 
     try { 
      bufferedReader.close(); 
     } catch(Exception e) { 
      e.printStackTrace(); 
     } 
    } 

    return content; 

} 

這將返回頁面的字符串表示形式。如果頁面包含HTML，它將返回text/html，如果它包含文本，它將只返回文本。

然後，作爲前一個用戶說，你可以設置你的GUI

protected void onPostExecute(String result) { 
    textView3.setText(result); 
} 

上面的示例將在技術上的工作響應。不過，我強烈推薦使用HttpURLConnection上的http://loopj.com/android-async-http/。你的代碼會更好，這將是重要的。

，我送你的代碼假設參數始終PIN = OFF1，因爲它更