Sending send data through URL parameters in Android

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)