Sending send data through URL parameters in Android - php

Question

I'm trying to make an app when users fill form, and click send button, to send informations through url parametrs like :
localhost/my.php?param1=Name&param2=SureName
I have read through internet about it, but can't get it, I found some examples but they tell how to pass through Json, I want just to send through link. (I think is easiest way )
My php code is:
<?php
require "connection.php";
if (!isset($_GET['param1'])) {
echo '<p align="center"> No data passed!</p>"';
}
if (strcmp($_GET['param1'],'FirstParam')== 0)
{
$sql= "Query HERE TO ADD INTO DB";
mysqli_query($connection,$sql);
if ($connection->query($sql) === TRUE) {
echo "Record updated successfully";
} else {
echo "Error updating record: " . $connection->error;
}
mysqli_close($connection);
}
Any help will be appreciated.

Answer 1

I recommend to have a look on OkHttp. It is an easy library to send HTTP requests in whatever way you like. On Vogella you can also find an example of sending a GET request with additional parameters. Basically it looks like this
HttpUrl.Builder urlBuilder = HttpUrl.parse("https://api.github.help").newBuilder();
urlBuilder.addQueryParameter("v", "1.0");
urlBuilder.addQueryParameter("user", "vogella");
String url = urlBuilder.build().toString();
Request request = new Request.Builder()
.url(url)
.build();
And after that you need to send your request (asynchronously) with the help of the OkHttpClient like this
client.newCall(request).enqueue(new Callback() {
#Override
public void onFailure(Call call, IOException e) {
e.printStackTrace();
}
#Override
public void onResponse(Call call, final Response response) throws IOException {
if (!response.isSuccessful()) {
throw new IOException("Unexpected code " + response);
} else {
// do something wih the result
}
}
(Code was taken from Vogella)