java - 发出 HTTPS 请求时出现问题

Question

我有这个异步任务:

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;
    }       
}

我正在尝试将参数传递给我的 php 脚本并更新数据库中的值。使用上面的代码，我的链接应该如下所示:https://site.com/android/jokes/updateLikes.php?jokeName=Joke 7

问题是请求之后没有任何反应。我知道这是正常的，因为 HTTPS，但我真的找不到一种方法来使请求正常工作，即使这里有大约 10 个答案。你能指导我并告诉我我缺少什么吗？我知道它非常小，但我无法发现它。

这是我的 php 代码(只是附加信息):

<?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.S我已在发布的代码中更改了网站名称和数据库详细信息。

Answer 1

您应该使用“url 编码”来确保目标字符串中的空白得到正确转换。