的Android - 發送HTTPS Get請求

Question

我想送一個HTTPS GET請求到谷歌購物API但是沒有什麼很爲我工作，例如這裏就是我想要的那一刻：

try {   
     HttpClient client = new DefaultHttpClient(); 
     HttpGet request = new HttpGet(); 
     request.setURI(new URI("https://www.googleapis.com/shopping/search/v1/public/products/?key={my_key}&country=&q=t-shirts&alt=json&rankByrelevancy=")); 
     HttpResponse response = client.execute(request); 
    } catch (URISyntaxException e) { 
     e.printStackTrace(); 
    } catch (ClientProtocolException e) { 
     // TODO Auto-generated catch block 
     e.printStackTrace(); 
    } catch (IOException e) { 
     // TODO Auto-generated catch block 
     e.printStackTrace(); 
    } 
    return response; 
}  

如果任何人有任何建議，如何改善或取代它請讓我知道，在此先感謝。

Answer 1

你應該會收到一個編譯錯誤。

這是正確的版本：

HttpResponse response = null; 
try {   
     HttpClient client = new DefaultHttpClient(); 
     HttpGet request = new HttpGet(); 
     request.setURI(new URI("https://www.googleapis.com/shopping/search/v1/public/products/?key={my_key}&country=&q=t-shirts&alt=json&rankByrelevancy=")); 
     response = client.execute(request); 
    } catch (URISyntaxException e) { 
     e.printStackTrace(); 
    } catch (ClientProtocolException e) { 
     // TODO Auto-generated catch block 
     e.printStackTrace(); 
    } catch (IOException e) { 
     // TODO Auto-generated catch block 
     e.printStackTrace(); 
    } 
    return response; 
} 

因此，如果現在你有你的迴應將返回爲空的錯誤。

一旦你有了迴應，並且檢查它爲空，你就會想要得到內容（即你的JSON）。

http://developer.android.com/reference/org/apache/http/HttpResponse.html http://developer.android.com/reference/org/apache/http/HttpEntity.html http://developer.android.com/reference/java/io/InputStream.html

response.getEntity().getContent(); 

這給你一個InputStream一起工作。如果你想將其轉換爲一個字符串，你會做以下或等價的：

http://www.mkyong.com/java/how-to-convert-inputstream-to-string-in-java/

public static String convertStreamToString(InputStream inputStream) throws IOException { 
     if (inputStream != null) { 
      Writer writer = new StringWriter(); 

      char[] buffer = new char[1024]; 
      try { 
       Reader reader = new BufferedReader(new InputStreamReader(inputStream, "UTF-8"),1024); 
       int n; 
       while ((n = reader.read(buffer)) != -1) { 
        writer.write(buffer, 0, n); 
       } 
      } finally { 
       inputStream.close(); 
      } 
      return writer.toString(); 
     } else { 
      return ""; 
     } 
    } 

當你有這樣的字符串，你需要從它創建一個JSONObject：

http://developer.android.com/reference/org/json/JSONObject.html

JSONObject json = new JSONObject(inputStreamAsString); 

完成！

Answer 2

你添加到您的清單

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

Answer 3

很難肯定地說，如果你不告訴我們是什麼錯誤。

但是，如果您在UI線程上運行此操作，並且Web服務器需要超過幾秒的響應時間，則系統會收到「應用程序無響應」警告。確保您在單獨的線程上進行任何網絡傳輸。

Answer 4

，您可以嘗試這種方式也許使用URLConnection類

String error = ""; // string field 
private String getDataFromUrl(String demoIdUrl) { 

    String result = null; 
    int resCode; 
    InputStream in; 
    try { 
     URL url = new URL(demoIdUrl); 
     URLConnection urlConn = url.openConnection(); 

     HttpsURLConnection httpsConn = (HttpsURLConnection) urlConn; 
     httpsConn.setAllowUserInteraction(false); 
     httpsConn.setInstanceFollowRedirects(true); 
     httpsConn.setRequestMethod("GET"); 
     httpsConn.connect(); 
     resCode = httpsConn.getResponseCode(); 

     if (resCode == HttpURLConnection.HTTP_OK) { 
      in = httpsConn.getInputStream(); 

      BufferedReader reader = new BufferedReader(new InputStreamReader(
        in, "iso-8859-1"), 8); 
      StringBuilder sb = new StringBuilder(); 
      String line; 
      while ((line = reader.readLine()) != null) { 
       sb.append(line).append("\n"); 
      } 
      in.close(); 
      result = sb.toString(); 
     } else { 
      error += resCode; 
     } 
    } catch (IOException e) { 
     e.printStackTrace(); 
    } 
    return result; 
}