AsyncTask 不返回任何响应

Posted 2023-05-07

【中文标题】AsyncTask 不返回任何响应

【英文标题】：AsyncTask not returning any response

【发布时间】：2017-12-08 18:27:17

【问题描述】：

您好，我无法从主机获取我的网址的响应。我尝试放置静态变量并且它有效，但是当我尝试使用 asynctask 进行响应时它不起作用。

<?php
include_once('connection.php');

$where='';

if(isset($_GET['userLat']) && isset($_GET['userLng']))

$where = "WHERE ((userLat >= '".addslashes($_GET['userLat'])."' AND userLat <= '".addslashes($_GET['userLat'])."' + .00901) OR (userLat <= '".addslashes($_GET['userLat'])."' AND userLat >= '".addslashes($_GET['userLat'])."' - .00901)) AND ((userLng >= '".addslashes($_GET['userLng'])."' AND userLng <= '".addslashes($_GET['userLng'])."' + .014935) OR (userLng <= '".addslashes($_GET['userLng'])."' AND userLng >= '".addslashes($_GET['userLng'])."' - .014935))";

    $query = "SELECT * FROM tbl_user ".$where." AND isOnline = 'Yes' LIMIT 1 ";
$result = mysqli_query($conn, $query);

$json = array();

if(mysqli_num_rows($result))
    while($row = mysqli_fetch_assoc($result))
     $json['details'][]=$row;


mysqli_close($conn);
echo json_encode($json);

?>

异步任务

    private class JsonTask extends AsyncTask<String, String, String> 

    protected void onPreExecute() 
        super.onPreExecute();

        pd = new ProgressDialog(getActivity());
        pd.setMessage("Checking nearby wingmans");
        pd.setCancelable(false);
        pd.show();
    

    protected String doInBackground(String... params) 


        HttpURLConnection connection = null;
        BufferedReader reader = null;

        try 
            URL url = new URL(params[0]);
            connection = (HttpURLConnection) url.openConnection();
            connection.connect();

            InputStream stream = connection.getInputStream();

            reader = new BufferedReader(new InputStreamReader(stream));

            StringBuffer buffer = new StringBuffer();
            String line = "";

            while ((line = reader.readLine()) != null) 
                buffer.append(line + "\n");
                Log.d("Response: ", "> " + line);   //here u ll get whole response...... :-)

            

            return buffer.toString();

         catch (MalformedURLException e) 
            e.printStackTrace();
         catch (IOException e) 
            e.printStackTrace();
         finally 
            if (connection != null) 
                connection.disconnect();
            
            try 
                if (reader != null) 
                    reader.close();
                
             catch (IOException e) 
                e.printStackTrace();
            
        
        return null;
    

    @Override
    protected void onPostExecute(String result) 
        super.onPostExecute(result);
        if (pd.isShowing()) 
            pd.dismiss();
        
        request = result;
        Log.d("Test", result);
    

我如何称呼我的网址

        btnEmergency.setOnClickListener(new View.OnClickListener() 
        @Override
        public void onClick(View v) 
            String url = "http://10.0.2.2/wingman/emergency.php?userLat="+tvLat.getText().toString()+"ANDuserLng="+tvLng.getText().toString();
            new JsonTask().execute(url);
        
    );

谢谢大家，我希望我能得到我的问题的答案。谢谢:)

【问题讨论】：

是否显示异常日志？

你真的应该发布日志，这可能有很多原因，首先想到的是互联网许可

@UsmanRana - 也没有异常日志

【参考方案1】：

我太傻了。我将我的网址中的AND 更改为& 来自

String url = "http://10.0.2.2/wingman/emergency.php?userLat="+tvLat.getText().toString()+"ANDuserLng="+tvLng.getText().toString();

到

String url = "http://10.0.2.2/wingman/emergency.php?userLat="+tvLat.getText().toString()+"&userLng="+tvLng.getText().toString();

它奏效了。谢谢！