sending values from android to php without namepair value - php

Question

I want to send data from android to php.Am using android api 23 and from android's developers site a got to know that namepairvalue is deprecated from api 22 and above. I referred to one of the post here in stackoverflow and got the below code. but still it doesn't work.
My problem is that am sending a value(ID) to php code and based on that value fetching the records from database. Now if i fetch data without using this value it works fine, but i am unable to send the id value to php code.
i am unable to find whats wrong in this code or may be my fault is editing the code. It would be great if someone helps me to understand it clearly .
Thank you in advance.
Here is the code snippet that i have tried.
Android Code:
String sid = staffid.toString(); // this sid is passed as intent from another activity and passing it to below link.
Log.d("SID :", "" + sid);
try {
URL url = null;
url = new URL("http://usplhubli.96.lt/hrm/hrm_app/individual_view_staff.php");
Log.d("url to display :", "" + url.toString());
HttpURLConnection conn = null;
conn = (HttpURLConnection) url.openConnection();
conn.setReadTimeout(CONNECTION_TIMEOUT);
conn.setConnectTimeout(CONNECTION_TIMEOUT);
conn.setRequestMethod("POST");
conn.setDoInput(true);
conn.setDoOutput(true);
DataOutputStream os = new DataOutputStream(conn.getOutputStream());
// Log.d("os to display :", "" + os.toString());
Uri.Builder builder = new Uri.Builder().appendQueryParameter("sid", sid);
Log.d("builder to display :", "" + builder.toString());
String query = builder.build().getEncodedQuery();
BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(os, "UTF-8"));
Log.d("writer display :", "" + writer.toString());
writer.write(query);
writer.flush();
writer.close();
os.close();
} catch (MalformedURLException e1) {
e1.printStackTrace();
} catch (ProtocolException e2) {
e2.printStackTrace();
} catch (IOException e3) {
e3.printStackTrace();
}
return null;
}
I have this code in doInBackground(String... params) of AsyncTask
based on this sid value i have to get the data from database and display it in android activity.
Here is the PHP code that i have tried
PHP Code:
$response=array();
$sid=$_POST["sid"]; // this is the POST data that am send from android
//$sid="4"; // this is the static value that i have tried and its working
print_r($sid);
include 'dbconfig.php';
$imagePath='http://www.usplhubli.96.lt/hrm';
$query=mysqli_query($conn,"select * from hrm_staff where sid='$sid'");
if(mysqli_num_rows($query) >0) {
$response["hrm_staff"] = array();
while ($row = mysqli_fetch_array($query)) {
$recp = array();
$recp["sid"]=$row["sid"];
$fullName=$row['fname']."".$row['mname']."".$row['lname'];
$recp["name"]= $fullName;
$recp["address"]=$row['address'];
$recp['city']=$row['city'];
$recp["design"]=$row['did'];
$recp["contact"]=$row['mobile'];
$recp["email"]=$row['email'];
$recp['qualification']=$row['quali'];
$recp["dateofjoining"]=$row['doj'];
$ppic1=$row['ppic'];
$ppic1;
$ppic21=$imagePath."/".trim($ppic1);
$recp["ppic"]= $ppic21;
array_push($response["hrm_staff"], $recp);
}
$response["success"] = 1;
$response["message"]="display records";
echo json_encode($response);
}
else
{
$response["success"] = 0;
$response["message"] = "No products found";
echo json_encode($response);
}
Please suggest. Thank you.

Answer 1

use these lines of code for using HttpURLConnection
For sending the request parameter are as:-
Uri.Builder builder = new Uri.Builder()
.appendQueryParameter("phone", number)
.appendQueryParameter("password", password)
.appendQueryParameter("device_login",value);
And for the getting the request parameter in the post method of connectionUrl are as follow:-
URL url = new URL(myurl);
HttpURLConnection urlConnection = (HttpURLConnection) url.openConnection();
urlConnection.setReadTimeout(30000);
urlConnection.setConnectTimeout(30000); ;
String query = builder.build().getEncodedQuery();
System.out.println("REQUEST PARAMETERS ARE===" + query);
urlConnection.setRequestMethod("POST");
urlConnection.setDoInput(true);
urlConnection.setDoOutput(true);
urlConnection.connect();
//Send request
DataOutputStream wr = new DataOutputStream (
urlConnection.getOutputStream ());
wr.writeBytes (query);
wr.flush ();
wr.close ();
//Get Response
InputStream isNew = urlConnection.getInputStream();
BufferedReader rd = new BufferedReader(new InputStreamReader(isNew));
String line;
response = new StringBuffer();
while((line = rd.readLine()) != null) {
response.append(line);
}
rd.close();
System.out.println("Final response===="+response);

Answer 2

There might be more than one issue but a big one is you never send the request, all you are doing is creating the request and setting up the parameters
you need to do a
conn.connect();
or
conn.getInputStream();
or any number of things you can do to send the request and get the information you require
AFTER you write your params