如何从Android的值传递给PHP的Web服务和检索吗？ [英] How to pass a value from android to php web service and retrieve it?

问题描述

我试图将值传递给我的PHP web服务。我已经使用这个code为通过名值：

I'm trying to pass a value to my php webservice. I already use this code for pass the "name" value :

 private class MyAsyncTask extends AsyncTask<String, Void, Void> {

    protected Void doInBackground (String... params)
    {
            Intent intent = getIntent();
            String name = intent.getStringExtra("KEY_NAME");
            //HttpClient httpclient = new DefaultHttpClient();
            HttpPost httppost = new HttpPost("http://10.0.2.2/secure_login/get_data_user.php");

            List<NameValuePair> nameValuePair = new ArrayList<NameValuePair>(1);
            nameValuePair.add(new BasicNameValuePair("KEY_NAME", name));
            DefaultHttpClient hc = new DefaultHttpClient();
           // HttpResponse response = hc.execute(httppost);


            try {
                httppost.setEntity(new UrlEncodedFormEntity(nameValuePair));
            } catch (UnsupportedEncodingException e){
                // writing error to log
                e.printStackTrace();
            }

            try {
                HttpResponse response = hc.execute(httppost);
                HttpEntity entity = response.getEntity();
                InputStream inStream = entity.getContent();
                // writing response to log
                Log.d("Http Response:", response.toString());

               } catch (ClientProtocolException e) {
                   // writing exception to log
                   e.printStackTrace();
               } catch (IOException e) {
                   e.printStackTrace();
               }
            return null;



    }

和这一点，流转换为字符串。

And this, for convert Stream to String.

          protected String convertStreamToString(InputStream inStream)
            {
               BufferedReader reader = new BufferedReader(new InputStreamReader(inStream));
               StringBuilder sb = new StringBuilder();

               String line = null;
               try 
               {
                   while ((line = reader.readLine()) != null) 
                   {
                       sb.append(line + "\n");
                   }
               } 
               catch (IOException e) 
               {
                   e.printStackTrace();
               } 
               finally 
               {
                   try 
                   {
                       inStream.close();
                   } 
                   catch (IOException e) 
                   {
                       e.printStackTrace();
                   }
               }
               return sb.toString();
            }
  }

但我只拿到了日志猫这样的响应：
org.apache.http.message.BasicHtt$p$psponse@43e4c068

But I only got this response in log cat : org.apache.http.message.BasicHttpResponse@43e4c068

我需要通过名称值，使我的PHP Web服务可以获取并做查询是这样的：

I need to pass the "name" value so my php webservice can retrieve and do the query like this :

 if (isset($_GET['name'])) {
$name = $_GET['name'];

require_once 'DB_Functions.php';
$db = new DB_Functions();
$result = mysql_query("SELECT name, email from users where name = '$name'");

我如何应该修复它？谢谢你在前进。

How do I should fix it? Thank you in advance.

推荐答案

您在Android的发送POST请求，并试图检索与 $ PHP中的参数_ GET 。

You are sending a post request in android and trying to retrieve the parameter in PHP with $_GET.

尝试通过 $ _ POST访问名称变量：

if (isset($_POST['name']))
    $name = mysql_real_escape_string($_POST['name']);

require_once 'DB_Functions.php';
$db = new DB_Functions();
$result = mysql_query("SELECT name, email from users where name = '$name'");

重要：不要忘记你把他们之前在SQL语句中使用 mysql_real_escape_string 来避免SQL注入逃避字符串

Important: Don't forget to escape the strings before you put them in your SQL statement with mysql_real_escape_string to avoid SQL injections.

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