从Android电子连接MySQL数据库数据库、电子、Android、MySQL

2023-09-06 00:57:55 作者：那年丶我们都狠霸道

嗯，这是在code片断我用来访问getUser.php以检索在我的应用程序从MySQL数据库用户的详细信息：

well this is the code snippet i use to access the getUser.php to retrive user details from a MySQL database in my application:

String result = "";

    //the year data to send

    ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();

    nameValuePairs.add(new BasicNameValuePair("uid","demo"));

         //http post
         try{
                 HttpClient httpclient = new DefaultHttpClient();

           HttpPost httppost = new HttpPost("http://192.xxx.xx.xxx/getUser.php");

            httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
                HttpResponse response = httpclient.execute(httppost);
                HttpEntity entity = response.getEntity();
                InputStream is = entity.getContent();
        }catch(Exception e){
                Log.e("log_tag", "Error in http connection "+e.toString());
        }
        //convert response to string
        try{
                InputStream is = null;
                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){
                Log.e("log_tag", "Error converting result "+e.toString());
        }

        //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("id")+
                              ", name: "+json_data.getString("fname")+
                                ", sex: "+json_data.getInt("sex")+
                                ", birthyear: "+json_data.getInt("dob")
                        );
                }

  }
        catch(JSONException e){
                Log.e("log_tag", "Error parsing data "+e.toString());
        }

}

本段摘自 http://helloandroid.com

所有的设定罚款：MySQL数据库，IIS FastCGI的，PHP工具和驱动程序。的http：// 192甚至从浏览器的URL时调用下面的脚本。 xxx.xx.x.xxx/getUser.php?uid=demo 工作正常，但是在android的是显示java.lang.NullPointerException 和 org.json.JSONEXCEPTION返回错误：输入的字符0结束

Everything is configured fine: the MySQL Db, IIS with FASTCGi, PHP tools and drivers. even the script below when called from browser with url: http://192.xxx.xx.x.xxx/getUser.php?uid=demo works fine, But returns error in android with java.lang.NullPointerException and org.json.JSONEXCEPTION: End of input at character 0

<?php
mysql_connect("myhost","username","pwd");
mysql_select_db("mydb");

$q=mysql_query("SELECT * FROM userinfo WHERE uid ='".$_REQUEST['uid']."'");

while($e=mysql_fetch_assoc($q))
        $output[]=$e;

print(json_encode($output));

mysql_close();
?>

在本节中任何人都可以帮忙吗？

Can anybody help in this section?

问候，米斯特里Hardik

Regards, Mistry Hardik

从Android电子连接MySQL数据库数据库、电子、Android、MySQL

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