我有一个异步任务：

public class likeTheJoke extends AsyncTask<Void, Integer, Void>{        

    @Override
    protected Void doInBackground(Void... params) {

        // Create a new HttpClient and Post Header
        HttpClient httpclient = new DefaultHttpClient();
        HttpPost httppost = new HttpPost("https://site.com/android/jokes/updateLikes.php");

        try {
             // Add data
            String encodedUrl = URLEncoder.encode("Joke 7", "UTF-8"); // recently added after a comment suggestion
            List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
            nameValuePairs.add(new BasicNameValuePair("jokeName", encodedUrl));
            httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

            // Execute HTTP Post Request
            HttpResponse response = httpclient.execute(httppost);

        } catch (ClientProtocolException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return null;
    }       
}

我正在尝试向php脚本传递一个参数，并更新数据库中的一个值。根据上面的代码，我的链接应该是这样的： https://site.com/android/jokes/updateLikes.php?jokeName=Joke 7 问题是在请求之后，什么都没有发生。我知道这很正常，因为 HTTPS ，但我真的找不到一种方法来让请求生效，即使这里有大约10个答案。你能告诉我我错过了什么吗？我知道这是一件很小的东西，但我看不出来。
下面是我的php代码（只是一个附加信息）：

<?php
$jokeName = urldecode($_GET["jokeName"]); // urldecode added after a comment suggestion.

$con=mysqli_connect("localhost","user","pass","database");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

mysqli_query($con,"UPDATE JOKES SET likes = likes+1
WHERE jokeName='$jokeName'");

mysqli_close($con);
?>

p、 我已经改变了网站名称和数据库的详细信息在张贴的代码。