Question

Here my code, I need to insert into mysql database I have mysql,php,android

1) The problem i have, i need a successful result to my php so that I can be able to insert using android.so my code for php:

2) The table I’m using has a foreign key : Provinces_idProvinces.

 $result = mysql_query("INSERT INTO property( Pname, P_Price,P_Desc,P_City, P_Size,P_Rooms, P_garage, P_Address, P_Long, P_Lat, Provinces_idProvinces)  
VALUES('$_POST[Pname]',$_POST[P_Price],'$_POST[P_Desc]','$_POST[P_City]','$_POST[P_Size]','$_POST[P_Rooms]',$_POST[P_garage],'$_POST[P_Address]',$_POST[P_Long],$_POST[P_Lat],$_POST[Provinces_idProvinces])");

 if ($result) {
        // successfully inserted into database
        $response["success"] = 1;
        $response["message"] = $result ;

        // echoing JSON response
        echo json_encode($response);
    } else {
        // failed to insert row
        $response["success"] = 0;
        $response["message"] = "Oops! An error occurred.";

        echo $response["success"];

        // echoing JSON response
        echo json_encode($response);
    }

And my class which I pass parameters but it giving me an error :

List<NameValuePair> params = new ArrayList<NameValuePair>();
          Log.d("Parameters","1");
          params.add(new BasicNameValuePair("Pname",Pname));
          params.add(new BasicNameValuePair("P_Price",Prices));        
          params.add(new BasicNameValuePair("P_Desc",Desc));
          params.add(new BasicNameValuePair("P_City",city));
          params.add(new BasicNameValuePair("P_Size",Size));
          params.add(new BasicNameValuePair("P_Rooms",bedNumber));
          params.add(new BasicNameValuePair("P_garage",Garnums));
          params.add(new BasicNameValuePair("P_Address",Address));
          params.add(new BasicNameValuePair("P_Long",longertude));
          params.add(new BasicNameValuePair("P_Lat",lattitude));
          params.add(new BasicNameValuePair("Provinces_idProvinces",Provnums));

          Log.d("Json","obj");
       // getting JSON Object
          // Note that create product url accepts POST method

          JSONObject json = jsonParser.makeHttpRequest(save_prop,
                  "POST", params);
          // check log cat fro response
          Log.d("Create Response", json.toString());

Which I’m getting an error in JSON because of the value 0, which is returned by php,
I DO NOT KNOW HOW TO SOLVE THIS PROBLEM

And here my logCat :

06-25 19:36:40.175: E/JSON Parser(2261): Error parsing data org.json.JSONException: Value 0 of type java.lang.Integer cannot be converted to JSONObject
06-25 19:36:40.175: W/dalvikvm(2261): threadid=10: thread exiting with uncaught exception (group=0x40014760)
06-25 19:36:40.236: E/AndroidRuntime(2261): FATAL EXCEPTION: AsyncTask #1
06-25 19:36:40.236: E/AndroidRuntime(2261): java.lang.RuntimeException: An error occured while executing doInBackground()
06-25 19:36:40.236: E/AndroidRuntime(2261):     at android.os.AsyncTask$3.done(AsyncTask.java:266)

Answer 1

EDIT: Simply just remove echo $response["success"];.

You shouldn’t print out the results like the way you have. I assume this is what the results would look like when you hit the server with that request…

0{"results":"..."}

You’re getting this error because the result (shown above) is being consumed as a JSON Object on the Android side. The results above IS NOT in a valid JSON format thus an exception is thrown. So my suggestion is to rework your PHP to return results that look like…

{"success":0,"results":"..."}