发出简单的HTTP请求并接收内容

Question

我发现了一个工作代码，用于在这里制作简单的HTTP请求，从How can I make a simple HTTP request in MainActivity.java? (Android Studio)开始，我将在下面发布它（有一些更改，如果我没有错，现在需要使用try{} catch{}）。但是我想问一下如何获得内容？我在下面的方式与代码打交道：

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)通话

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

然后填写你的文本字段或任何其它数据都需要更新。

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

Answer 2

你到底在哪里卡住了？您的代码大部分都是正确的，但您可能需要稍微重新排列它。你的范围是有点关闭，你的评论代码几乎到达那里。见下面

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; 

} 

这将返回页面的字符串表示形式。如果页面包含HTML，它将返回text/html，如果它包含文本，它将只返回文本。

然后，作为前一个用户说，你可以设置你的GUI

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

上面的示例将在技术上的工作响应。不过，我强烈推荐使用HttpURLConnection上的http://loopj.com/android-async-http/。你的代码会更好，这将是重要的。

，我送你的代码假设参数始终PIN = OFF1，因为它更