phonegap/msql(i) can't establish connection to database - mysql

Question

i'm currently developing an web app for iOS but the app can't get the connection right.
when i'm trying to log in, i get the following iOS popup: index.html Error.
while the PHP files are online,
here some of my code:
connection.php
<?php //Make connection with database
// Verbinden met MySQL Database
$host = "localhost"; // Welke server : localhost
$username = "*******"; // Gebruikersnaam
$password = "****"; // Wachtwoord
$dbnaam = "*****"; // Naam van de database
$db_error1 = "<p>FOUT: verbinden met databaseserver is mislukt</p>"; // Foutmelding 1
$db_error2 = "<p>FOUT: selecteren van database is mislukt</p>"; // Foutmelding 2
$db_error3 = "<p>FOUT: sluiten van database is mislukt</p>"; // Foutmelding 3
// Verbinden met Databaseserver
$con=mysqli_connect($host, $username, $password, $dbnaam);// or die($db_error1);
// verbinden met de database
//mysql_select_db($dbnaam, $db) or die($db_error2);
if (mysqli_connect_errno($con))
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
?>
demo.js
// define the URL of the server component
//var url ="http://www.jorisgraaumans.nl/dutchmobile/";
//var url = "http://localhost:8888/Mobile/Festival/";
var url = "http://www.rikvandoorn.com/mobile/";
// afvangen van het standaard submit event
// zie ook: http://api.jquery.com/submit/
$('#login').on('pageinit', function(event) {
$('#loginForm').submit(function() {
$.ajax({
type: "POST",
url: "processLogin.php",
cache: "false",
dataType: "json",
data: {
email : $('#email').val(),
wachtwoord : $('#wachtwoord').val(),
},
success: function(phpData){
$("#return").data("login", phpData.login);
console.log("login is: "+ phpData.login);
if(phpData['error'] == true){
$.mobile.changePage("#return", {
transition : "fade"
})
} else {
$.mobile.changePage("#home", {
transition : "fade"
})
}
},
error: function(){
alert('Error');
}
});
return false; // return false to prevent the default submit of the form to the server.
});
// end: pageinit loginForm
});
$('#return').on('pageshow', function(event) {
$("#error_message_log").empty();
$("#error_message_log").prepend('<p>' + $(this).data('login') + '</p>');
});
processLogin.php
<?php //process login form
include 'http://www.rikvandoorn.nl/mobile/connect.php'; //connection to database
if(empty($_POST['email']) || empty($_POST['wachtwoord'])){
$return['error'] = true; //return error
$return['login'] = 'Niet alle velden zijn ingevuld';
} else {
$email = $_POST['email'];
$wachtwoord = $_POST['wachtwoord'];
//Check if user exists
$sql = "SELECT * FROM gebruiker WHERE email = '$email'";
$result = mysql_query($sql) or die(mysql_error());
$num_rows = mysql_num_rows($result);
if($num_rows == 0){
$return['error'] = true;
$return['login'] = 'Gebruiker bestaat niet';
} else {
//Check if password is correct for user
$sql_2 = "SELECT * FROM gebruiker WHERE email = '$email' && wachtwoord = '$wachtwoord'";
$query_2 = mysql_query($sql_2);
$num_rows_2 = mysql_num_rows($query_2);
if($num_rows_2 == 0){
$return['error'] = true;
$return['login'] = 'Wachtwoord onjuist';
} else {
$return['error'] = false;
$return['login']['email'] = $email;
$return['login']['wachtwoord'] = $wachtwoord;
}
}
}
echo json_encode($return);

Answer 1

Unless you've riddled the code with typos, you're mixing two entirely seperate mysql libraries:
$con=mysqli_connect($host, $username, $password, $dbnaam);// or die($db_error1);
^---note the presence of an 'i'
and in the other scripts:
$result = mysql_query($sql) or die(mysql_error());
^---note the LACK of an 'i'
mysql and mysqli are NOT interchangeable, and connections established in one are utterly useless for the other.
The mysql (WITHOUT an i) library is obsolete and deprecated. Stick with mysqli (WITH an i) only.