Sign Up button does not work correctly

[AlJohri]

Clicking the "Sign Up" button on your landing page returns the php submission code with your database password. Please fix this.

You can simulate this with the following command:

curl -X POST http://transad.x10host.com/submit1.php --data "first_name=Al&last_name=Johri&email=al.johri@gmail.com"

Returns:

<?php
define('DB_NAME', 'table_input');
define('DB_USER', 'root');
define('DB_PASSWORD', 'Aakash');
define('DB_HOST', 'localhost');

$link = mysql_connect("localhost",DB_USER,DB_PASSWORD);

if (!$link) {
        die('Could not connect: ' . mysql_error());
}

$db_selected = mysql_select_db(DB_NAME, $link);

if (!$db_selected) {
        die('Can\'t use ' . DB_NAME . ': ' . mysql_error());
}

$first = $_POST['first_name'];
$last = $_POST['last_name'];
$email1 = $_POST['email'];

$sql = mysql_query("INSERT INTO user_input (first_name, last_name, email) VALUES ('$first', '$last', '$email1' )");

if (!$sql) {
        die('Error:'. mysql_error());
}

?>

[aakashmishra]

Hey, I am working on it there is a problem with the hosting site and I had a conversation with the TA regarding this on friday. The database works when used on my laptop server which I did show to him on Friday. Regards, Aakash Anand Mishra

On Jan 19, 2015, at 2:22 PM, Al Johri notifications@github.com wrote:

Clicking the "Sign Up" button on your landing page http://transad.x10host.com/register.html returns the php submission code with your database password. Please fix this.

You can simulate this with the following command:

curl -X POST http://transad.x10host.com/submit1.php --data "first_name=Al&last_name=Johri&email=al.johri@gmail.com" Returns:

<?php define('DB_NAME', 'table_input'); define('DB_USER', 'root'); define('DB_PASSWORD', 'Aakash'); define('DB_HOST', 'localhost');

$link = mysql_connect("localhost",DB_USER,DB_PASSWORD);

if (!$link) { die('Could not connect: ' . mysql_error()); }

$db_selected = mysql_select_db(DB_NAME, $link);

if (!$db_selected) { die('Can\'t use ' . DB_NAME . ': ' . mysql_error()); }

$first = $_POST['first_name']; $last = $_POST['last_name']; $email1 = $_POST['email'];

$sql = mysql_query("INSERT INTO user_input (first_name, last_name, email) VALUES ('$first', '$last', '$email1' )");

if (!$sql) { die('Error:'. mysql_error()); }

?> — Reply to this email directly or view it on GitHub https://github.com/nuvention-web/Transad/issues/3.

[AlJohri]

Sounds good.