【问题标题】:null value returned from the connection between android and MYSQL databaseandroid与MYSQL数据库连接返回空值
【发布时间】:2012-04-19 00:27:28
【问题描述】:

我正在尝试通过从 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。

当我运行应用程序时,由于返回的结果值为空而出现解析数据异常的错误?

为什么结果为空?错误是来自 PHP 脚本还是来自我的 java 代码?

请帮帮我

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();
        ?>

块引用

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());
            }


                     }
                                                                  }

                                                    }

【问题讨论】:

  • 听你在 mysql 中告诉我,City_Population 正在检索或多个数据

标签: php android json


【解决方案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();
         ?>

你试试这个确定它会起作用。

【讨论】:

    【解决方案2】:

    a) 您的脚本容易出现sql injections。在将 $_REQUEST[...] 参数放入 sql 查询字符串之前,您需要对其进行正确编码。
    b)您需要一些错误处理。任何 mysql_* 函数都可能失败,您的脚本必须处理这些错误情况。由于客户端需要一些 json 数据,因此只需将错误消息/代码也作为 json 编码数组返回。
    c) 您可能希望将Content-type 标头设置为application/json,请参阅RFC 4627http://docs.php.net/function.header

    <?php
    define('DEBUG_DETAILS', true);
    function onError($msg, $details) {
        $msg = array(
            'status'=>'error',
            'message'=>$msg
        );
        if ( defined('DEBUG_DETAILS') && DEBUG_DETAILS ) {
            $msg['details'] = $details;
        }
        die(json_encode($msg));
    }
    
    
    $mysql = mysql_connect("localhost","username","password") or OnError('database connection failed', mysql_error());
    mysql_select_db("Countries", $mysql) or OnError('database selection failed', mysql_error($mysql));
    
    $query = "
        SELECT
            City_Population
        FROM
            City
        WHERE
            Name='%s'
    ";
    $query = sprintf($query, mysql_real_escape_string($_REQUEST['Name'], $mysql));
    $sql=mysql_query($query, $mysql) or OnError('query failed', array('query'=>$query, 'errstr'=>mysql_error($mysql)));
    
    $output = array(
        'count'=>0,
        'records'=>array()
    );
    while( $row=mysql_fetch_assoc($sql) ) {
        $output['records'][]=$row;
        $output['count']+=1;
    }
    echo json_encode(array(
        'status'=>'ok',
        'result'=>$output
    ));
    

    您的 android 客户端应该收到一个对象字面量,例如

    {
      status:"ok",
      result: {
        'count': 2,
        'records': [ 10000, 15000]
      }
    }
    

    {
      status:"error",
      message: "database connection failed",
      setails: "...."
    }
    

    【讨论】:

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