Post variable via ajax to mysql db - phonegap - php

Question

I just want to post a simple variable from my phonegap app to mysql database. But it dont work. The db show me empty results.
Mysql database looks like this, Tablename is user
Rows:
user (varchar 25),vip (varchar 1),call (varchar 1)
At top of my project i inserted:
<script type="text/javascript" src="js/tcPlugin.js"></script>
<script type="text/javascript" src="js/phoneApp.js"></script>
<script type="text/javascript" src="js/start.js"></script>
start.js is my file for the ajax request - the other two working fine so i guess there is not the problem
my start.js file looks like this:
$(document).ready( function() {
var email = "john#web.de";
$.ajax({
type: 'post',
url: 'http://www.blablabla.de/phone/action.php',
data: {
data: {"email" : email },
//data:email,
},
success: function(result) {
console.log(result);
}
});
and my action.php looks like this:
<?php
$dbhost = "bla";
$dbuser = "bla";
$dbpass = "bla";
$dbname = "bla";
$user = $_POST['data']['email'];
$conn = new mysqli($dbhost, $dbuser, $dbpass, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO user (email)
VALUES ('" . $user . "')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
But everytime i start the app the db is empty. debugger concole is not giving me any errors. db connection is working, tried it with echo connection result. and the other js are working fine.

Answer 1

action.php
<?php
$dbhost = "bla";
$dbuser = "bla";
$dbpass = "bla";
$dbname = "bla";
print_r($_POST); // see full contents of the POST
$user = $_POST['data']['email'];
print PHP_EOL . $user . PHP_EOL; // see full contents of the $user var
$conn = new mysqli($dbhost, $dbuser, $dbpass, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO user (email) VALUES ('" . $user . "')";
print PHP_EOL . $sql . PHP_EOL; // see the actual sql query, you can test this later in your sql editor ex. phpmyadmin
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
make these edits, run the ajax and check the response tab in the debugger.