php - android与MYSQL数据库连接返回null值

Question

我试图通过从 Android 传递参数值来从我的 MySQL 数据库中检索特定数据，然后在查询的 PHP 脚本中读取该值以返回数据。

我运行应用程序时，由于返回的结果值为null，出现解析数据异常的错误？

为什么结果为空？错误是来自 PHP 脚本还是来 self 的 java 代码？

请帮帮我

提前致谢!

城市.php:

  <?php
     mysql_connect("localhost","username","password");
     mysql_select_db("Countries");
     $sql=mysql_query("select  City_Population  from City where Name= "'.$_REQUEST['Name']."'");
     while($row=mysql_fetch_assoc($sql))
     $output[]=$row;
      print(json_encode($output));
      mysql_close();
        ?>

Blockquote

Java 类:

       public class ConnectActivity extends ListActivity {

           String add="http://10.0.2.2/city.php";
           public void onCreate(Bundle savedInstanceState) {
           super.onCreate(savedInstanceState);
           setContentView(R.layout.main);

            new Connect().execute();

         }

  private class Connect extends AsyncTask<Void,Void,String>
   {     
             private  String result = "";
             private  InputStream is=null;
            private  String city_name="London";
           protected String doInBackground(Void... params) {
            try
          {
                  ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
                 nameValuePairs.add(new BasicNameValuePair("Name",city_name));
                 HttpClient httpclient = new DefaultHttpClient();
                HttpPost httppost = new HttpPost(add);
                httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
               HttpResponse response = httpclient.execute(httppost);
               HttpEntity entity = response.getEntity();
              is = entity.getContent();
                 }
        catch(Exception e)
           {
               Log.e("log_tag", "Error in http connection "+e.toString());
                 }


           //convert response to string
    try{
    BufferedReader reader = new BufferedReader(new InputStreamReader(is,"utf-8"),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){
           Log.e("log_tag", "Error converting result "+e.toString());
               }


          return result;
          }
       protected  void onPostExecute(String  result){

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

              TextView City_Name =(TextView)findViewById(R.id.city_name);
                                                                           TextView  City_population=(TextView)findViewById(R.id.city_pop);
                            City_Name.setText(json_data.getString(city_name));
                                                                          City_population.setText(population+"  " );
            }
            }
            catch(JSONException e){
            Log.e("log_tag", "Error parsing data "+e.toString());
            }


                     }
                                                                  }

                                                    }

Answer 1

     <?php
         $name=$_POST['NAME'];               
         mysql_connect("localhost","username","password");
         mysql_select_db("Countries");
         $sql=mysql_query("select  City_Population as citypop  from City where Name='$name' ");
         while($row=mysql_fetch_assoc($sql))
          $output=$row['citypop'];
        print(json_encode($output));
         mysql_close();
         ?>

你试试这个肯定会成功。