Reputation: 45
I have code for send parameters to php and get result in JSON :
private void details_location(int id){
HashMap<String, String> parameter = new HashMap<>();
parameter.put("id", id);
JsonObjectRequest jsonObjReq = new JsonObjectRequest(Method.GET, url, new JSONObject(parameter), new Response.Listener<JSONObject>() {
@Override
public void onResponse(JSONObject response) {
Log.d(TAG, response.toString());
try {
JSONArray respon=(JSONArray)response.get("location");
for (int i = 0; i < respon.length(); i++) {
JSONObject person = (JSONObject) respon.get(i);
id_location= person.getString("id_location");
address= person.getString("address");
}
} catch (JSONException e) {
e.printStackTrace();
Toast.makeText(getApplicationContext(),
"Error: " + e.getMessage(),
Toast.LENGTH_LONG).show();
}
}
}, new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
VolleyLog.d(TAG, "Error: " + error.getMessage());
Toast.makeText(getApplicationContext(),
error.getMessage(), Toast.LENGTH_SHORT).show();
}
});
}
and I have file php for proccessing query Mysql and have output in JSON :
<?php
require_once('db_connect.php');
if(isset($_GET['id'])){
$id=$_GET['id'];
$action="SELECT * FROM locaions WHERE id_location=\"$id\";";
$query=mysqli_query($db_connect, $action);
if (mysqli_num_rows($query) > 0)
{
$json['location']=array();
while($row=mysqli_fetch_assoc($query)){
$data=array();
$data["id_location"]=$row["id_location"];
$data["address"]=$row["address"];
array_push($json['location'], $data);
}
}
mysqli_close($db_connect);
echo json_encode($json);
}
else{
$action="SELECT * FROM locaions WHERE id_location=4;";
$query=mysqli_query($db_connect, $action);
if (mysqli_num_rows($query) > 0)
{
$json['location']=array();
while($row=mysqli_fetch_assoc($query)){
$data=array();
$data["id_location"]=$row["id_location"];
$data["address"]=$row["address"];
array_push($json['location'], $data);
}
}
mysqli_close($db_connect);
echo json_encode($json);
}
?>
when I run the program, the result always show location with id_location=4
(statement else). Why isset($_GET['id']) = false
? How to send parameter to PHP and get result in JSON ?
Upvotes: 0
Views: 2032
Reputation: 1422
This is the code for sending parameters to ur php server
protected String doInBackground(String... params) {
// TODO Auto-generated method stub
try {
String parameter1 = (String) params[0];
String parameter2 = (String) params[1];
String link = "Mention ur url link here/parameter1+parameter1 +parameter2+parameter2+";
URL url = new URL(link);
HttpClient client = new DefaultHttpClient();
HttpGet request = new HttpGet();
request.setURI(new URI(link));
HttpResponse response = client.execute(request);
BufferedReader in = new BufferedReader(new InputStreamReader(
response.getEntity().getContent()));
StringBuffer sb = new StringBuffer("");
String line = "";
while ((line = in.readLine()) != null) {
sb.append(line);
break;
}
in.close();
return sb.toString();
}
catch (Exception e) {
return new String("Exception: " + e.getMessage());
}
}
On post Execute the php should return data in the json format and you can handle this data in post Execute
Upvotes: 2