php– 从android和MYSQL数据库之间的连接返回的null值

概述我试图通过从Android传递参数的值从MySQL数据库中检索特定数据,然后在查询中的PHP脚本中读取此值以返回数据.当我运行应用程序时,出现错误解析数据异常,因为返回的结果值为null？为什么结果为null？是来自PHP脚本还是来自我的java代码的错误？请帮我提前致谢！city.php：<?php

我试图通过从Android传递参数的值从MysqL数据库中检索特定数据,然后在查询中的PHP脚本中读取此值以返回数据.

当我运行应用程序时,出现错误解析数据异常,因为返回的结果值为null？

为什么结果为null？是来自PHP脚本还是来自我的java代码的错误？

请帮我

提前致谢！

city.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());            }                     }                                                                  }                                                    }

解决方法:

     <?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();         ?>

你试着确定它会起作用.

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