Input in database failure - php

Question

I'm having trouble with the following code:
$sql= "INSERT INTO Users(Username, Password, Lastname, Email) VALUES
('$hash', '$lastname', '$email', '$email')";
mysqli_query($MyConnection, $sql);
if(!mysqli_query($MyConnection, $sql)) {
echo 'We are sorry, there are some problems with saving your data. Please try again within a few minutes.';
}
else {
echo 'We have succesfully saved your data. An activation e-mail will now be send to the e-mail address that you
have provided us.';
}
I get no direct errors as due to mistyping or misusing a function. I do get however the message of the if-statement in a failure, the "We are sorry(..)" text.
There must be a problem with the execution of the mysqli_query($MyConnection, $sql) function. But I don't see where it is.
P.S. I can't post images, because my reputation is below 10. (Which is quite weird to limit it to that point)
As some of you have provided most / all of the code:
<?php
// Opens the connection of the MySQL Database
$MyConnection = mysqli_connect('fdb6.biz.nf', '1446018_amp', '-')
or die("Could not connect to the database, please try again");
mysqli_select_db($MyConnection,'Users');
mysqli_connect_errno();
// Website Url:
$website = 'http://www.askmephilosophy.co.nf/';
// Information provided by the user
$username = $_POST['username'];
$password = $_POST['password']; // Will get encrypted.
$lastname = $_POST['lastname'];
$email = $_POST['email'];
// A higher "cost" is more secure but consumes more processing power
$cost = 5;
// Create a random salt
$salt = strtr(base64_encode(mcrypt_create_iv(16, MCRYPT_DEV_URANDOM)), '+', '.');
// Prefix information about the hash so PHP knows how to verify it later.
// "$2a$" Means we're using the Blowfish algorithm. The following two digits are the cost parameter.
$salt = sprintf("$2a$%02d$", $cost) . $salt;
// Hash the password with the salt
$hash = crypt($password, $salt);
$sql= "INSERT INTO Users(Username, Password, Lastname, Email) VALUES
('$hash', '$lastname', '$email', '$email')";
mysqli_query($MyConnection, $sql);
var_dump(mysqli_error($MyConnection));
if(mysqli_query($MyConnection, $sql)) {
echo 'We have succesfully saved your data. An activation e-mail will now be send to the e-mail address that you
have provided us.';
}
else {
echo 'We are sorry, there are some problems with saving your data. Please try again within a few minutes.';
mysqli_error($MyConnection);
}
mysqli_close($MyConnection);
?>

Answer 1

$sql= "INSERT INTO Users(Username, Password, Lastname, Email) VALUES
('$hash', '$lastname', '$email')";
This is your first issue; your table has four columns, and you're passing it three. This query is guaranteed to fail.
mysqli_query($MyConnection, $sql);
if(!mysqli_query($MyConnection, $sql)) {
You're calling the query function twice. You can do this with a single call:
if(!mysqli_query($MyConnection, $sql)) {
// add some error handling code here
// store the return value of mysqli_error() somewhere
echo 'We are sorry, there ar....';
Since you're using mysqli_, you should also be using prepared statements; I hope at least you're sanitising the database inputs before you try to add them to the database.

Answer 2

Why do you only have 3 values, it doesn't match the number of items you are trying to Insert (4) ...
$sql= "INSERT INTO Users(Username, Password, Lastname, Email) VALUES
('$username', '$hash', '$lastname', '$email')";
EDIT:
I would probably write it like this
$sql= "INSERT INTO Users(Username, Password, Lastname, Email) VALUES
({$username}, {$hash}, {$lastname}, {$email})";
EDIT:
Your password cannot be '-'
I would update your connection info like so:
$db = new mysqli('fdb6.biz.nf', 'user', 'pass', 'Users');
if($db->connect_errno > 0){
die('Unable to connect to database [' . $db->connect_error . ']');
}
EDIT AGAIN:
$myConnection = new mysqli('fdb6.biz.nf', 'user', 'pass', '1446018_amp');
$myConnection->mysqli_select_db($MyConnection,'Users');

Answer 3

try adding, I think you forgot this. Values always have to equal to columns
$sql= "INSERT INTO Users(Username, Password, Lastname, Email) VALUES
('$username', '$hash', '$lastname', '$email')";

Answer 4

First of all you are inserting twice that records, as there are two instances of mysqli_query($MyConnection, $sql);. You can just remove the first.
The problem here is that you are inserting 3 values in 4 fields.
Anyway you can get the specific error with
mysqli_error($MyConnection);
Add it at the end your echo forever or var_dump(mysqli_error($MyConnection)); in a new line.