从PHP / MySQL在Android上操作JSON对象

Question

这段代码可以正常工作...但是我不知道如何处理得到的结果。 在这里，我只是通过简单的查询从应用程序发送到网站

<?php
$con = mysql_connect("","user","pass") or die(mysql_error());
$objDB = mysql_select_db("tablename", $con) or die(mysql_error()); 

$username = mysql_real_escape_string($_POST['user1']); 
$sql = mysql_query("SELECT * FROM doctor WHERE username = '$username' ");

 while($row=mysql_fetch_assoc($sql)) 
 $output[]=$row;
  print(json_encode($output));
  mysql_close();
 ?>      

这是我的Android代码：

    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);
        //Saved preferences
        app_preferences = PreferenceManager.getDefaultSharedPreferences(this);
        button = (Button)findViewById(R.id.Button01);
        username= (EditText)findViewById(R.id.EditText01);//et
        password= (EditText)findViewById(R.id.EditText02);//et
        echoresponse= (TextView)findViewById(R.id.tv);//tv

     //   String str_user = app_preferences.getString("username", "0");
     //   String str_pass = app_preferences.getString("password", "0");
     //   String str_check = app_preferences.getString("checked", "0");

     //   if(str_check.equals("yes"))
     //   {
     //     username.setText(str_user);
     //     //password.setText(str_pass); 
     //   }

        button.setOnClickListener(new OnClickListener() {
            @Override
            public void onClick(View v) {
                try{
                    httpclient=new DefaultHttpClient();
                    httppost= new HttpPost("http://www.baemer.com/androiddealer.php"); // make sure the url is correct.
                    //add your data

                    nameValuePairs = new ArrayList<NameValuePair>(2);
                    // Always use the same variable name for posting i.e the android side variable name and php side variable name should be similar,
                    nameValuePairs.add(new BasicNameValuePair("user1",username.getText().toString().trim()));  // $Edittext_value = $_POST['Edittext_value'];
                    nameValuePairs.add(new BasicNameValuePair("pass1",password.getText().toString().trim()));  // $Edittext_value = $_POST['Edittext_value'];                   

                    httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
                    //Execute HTTP Post Request
                    response=httpclient.execute(httppost);
                    // edited by James from coderzheaven.. from here....
                    ResponseHandler<String> responseHandler = new BasicResponseHandler();
                    String res = httpclient.execute(httppost, responseHandler);
                    System.out.println("Response : " + res);
                    echoresponse.setText("Response from PHP : " + res);

                    HttpEntity entity = response.getEntity();
                    is = entity.getContent();
                }catch(Exception e){
                    System.out.println("Exception : " + e.getMessage());
                }

               //convert response to string

                try{

                    BufferedReader reader = new BufferedReader(new InputStreamReader(is, "iso-8859-1"),8);
                    StringBuilder sb = new StringBuilder();
                    String line = null;
                    while((line = reader.readLine())!=null){
                        sb.append(line+"\n");
                    }
                    is.close();
                    result=sb.toString();
                }catch(Exception e){
                    System.out.println("Exception : " + e.getMessage());//if not try Log. from http://www.gaanza.com/blog/how-to-mysql/
                }

                //parse JSON data

                try{
                    JSONArray jArray = new JSONArray(result);
                    for(int i =0;i<jArray.length();i++){
                        JSONObject json_data = jArray.getJSONObject(i);

                        Log.i("log_tag","id: "+json_data.getInt("dr_serial")+
                                ", username: "+json_data.getString("username"));

                    }
                }catch(JSONException e){
                    System.out.println("Exception : " + e.getMessage());//if not try Log. from http://www.gaanza.com/blog/how-to-mysql/
                }


            }


        });
    }
}

但是现在，当我按下按钮时，显示所有表都在Log.i（..... blabla）中说了。 如何操作这些列？ 如例如：分配一些变量dr_serial，然后将其显示在文本视图或其他内容上。

Answer 1

刚打电话

int dr_serial = json_data.getInt("dr_serial");

然后，您可以使用它做任何您想做的事情。