从Android应用程序连接和更新php服务器

我已经创建了一个gps应用程序，运行良好。现在我尝试的是将它连接到一个带有mysql数据库的php服务器上，其中所有人员的位置都会自动更新。。我是这部分的新用户。3.我试过一些，但我认为它写不出来。。有人能帮助我并指导我如何做的过程吗。。我们将非常感谢你的帮助。下面是我从frieds帮助和php脚本中编写的代码

public class Post extends LocService {{


TelephonyManager telephonyManager = (TelephonyManager)context.getSystemService(Context.TELEPHONY_SERVICE);
String deviceid = telephonyManager.getDeviceId();



    //this is JSON part to put your information inside it
    String postData = "{\"request\":{\"type\":\"locationinfo\"},\"userinfo\":{\"latitude\":\""+latitude+"\",\"longitude\":\""+longitude+"\",\"deviceid\":\""+deviceid+"\"}}";


    HttpClient httpClient = new DefaultHttpClient();

    // Post method to send data to server
    HttpPost post = new HttpPost("http://location.net/storeg.php");


    SQLiteDatabase db = databasehelper.getWritableDatabase();
    Cursor cursor = db.query(TABLE, null, null, null, null);

    cursor.moveToFirst();
    while(cursor.isAfterLast() == false) {

        if(cursor.getString(cursor.getColumnIndex("Sync")).equals("yes") ) {

            String mob = cursor.getString(cursor.getColumnIndex("MobileID"));
            String latitude = cursor.getString(cursor.getColumnIndex("Latitude"));
            String longitude = cursor.getString(cursor.getColumnIndex("Longitude"));
            String service = cursor.getString(cursor.getColumnIndex("Service"));


            JSONObject json = new JSONObject();
            try {
                json.put("MobileID", mob);
                json.put("Latitude", latitude);
                json.put("Longitude", longitude);
                json.put("Service", service);

            } catch (JSONException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
            try {
                receive = HttpPostExample.SendJsonUpdate(json, Sync_URL);
            } catch (ClientProtocolException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            } catch (IOException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
            Toast.makeText(context,  receive,Toast.LENGTH_SHORT).show();
        }
        cursor.moveToNext();    
    }
    cursor.close();













    try {
        post.setURI(new URI("http://location.net/storeg.php"));
    } catch (URISyntaxException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }

    // set your post data inside post method    
    try {
        post.setEntity(new StringEntity(postData));
    } catch (UnsupportedEncodingException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }

    // execute post request here 
    try {
        HttpResponse response = httpClient.execute(post);
    } catch (ClientProtocolException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    } catch (IOException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }



}
}

PHP脚本

<?php

$con = mysql_connect("localhost","root","");
if (!$con)
 {
  die('Could not connect: ' . mysql_error());
 }

mysql_select_db("mel_db", $con);

$latitude = $_POST['latitude'];
$longitude = $_POST['longitude'];
$service = $_POST['service'];
$devid = $_POST['devid'];



$sql = "INSERT INTO  `mehul_db`.`locations` (
`id` ,
`devid` ,
`latitude` ,
`longitude` ,
`service`
)
VALUES (
NULL ,  '$devid',  '$latitude',  '$longitude',  '$service'
);";
if (!mysql_query($sql,$con))
{
   die('Error: ' . mysql_error());
  }

mysql_close($con);

?>

请帮我完成这个过程。。。或者建议我该怎么做

*已编辑：*获取此代码更改变量后，我可以使用它获取纬度和经度吗

String result;
    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("name")+


                            ", sex: "+json_data.getInt("sex")+


                            ", birthyear: "+json_data.getInt("birthyear")


                   );


            }


   finally }


    }catch(JSONException e){


            Log.e("log_tag", "Error parsing data "+e.toString());


    }


}

---

您可以使用此代码通过post将参数发送到php，作为响应，您将获得inputstream。

您面临哪些问题？你遇到错误了吗？实际上我刚刚写了代码。。写作部分正确吗。。？你能告诉我怎么做吗..receive=HttpPostExample.SendJsonUpdatejson，Sync_URL；上面一行和这两行出错，SQLiteDatabase db=databasehelper.getwriteabledatabase；Cursor Cursor=db.queryTABLE，null，null，null，null；我们很难检查整个代码并验证它。你为什么不运行它，看看它是否有效，如果无效，带着你面临的问题回来。谢谢你。。我就试试看。。还有一件事我需要用json写东西吗？不，你不需要用json写东西。从数据库中获取latitude和longitude的值后，只需将其与代码中的键一起发送。nameValuePairs列表是键-值对。此处左边的值是键，在php中使用相同的键检索值。很抱歉再次打扰您..我将代码放在何处。。。？在什么之间。。？

     String url="http://location.net/storeg.php";

    DefaultHttpClient httpClient = new DefaultHttpClient();
    HttpPost httpPost = new HttpPost(url);

    List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(4);
    nameValuePairs.add(new BasicNameValuePair("latitude", latitude));
    nameValuePairs.add(new BasicNameValuePair("longitude", longitude));
    nameValuePairs.add(new BasicNameValuePair("service", service));
    nameValuePairs.add(new BasicNameValuePair("devid", devid));

    httpPost.setEntity(new UrlEncodedFormEntity(nameValuePairs));


    HttpResponse httpResponse = httpClient.execute(httpPost);
    InputStream httpEntity = httpResponse.getEntity().getContent();