PHP/MySQL - Data Not Being Stored In DB

[Matthew_16]

Note: this project is being restarted after one year of in-action, using the most relevant files I am able to locate. All final code will be adapted to MySQLi.

Versions of this code functioned properly to the best of my knowlege; now, with a new host I am having issues saving data to the db after a connection is made.

I have been researching the potential issue online but would like some advice regarding possible problems/errors regarding this issue before I proceed with editing the files.

Thank you in advance,
Matty

snippet of index.html
`

<?php

function NewUser()
{  
    $userName = $_POST['user'];
    $email = $_POST['email'];
    $password =  $_POST['pass'];
    $countries = $_POST['countries'];
    $state = $_POST ['state'];
    $city =  $_POST['city'];
    $company = $_POST ['company'];
    $manager1 =  $_POST['manager1'];
    $manager2 =  $_POST['manager2'];

    $query = "INSERT INTO X1 Table 1 (userName,email,password,countries,state,city,company,manager1,manager2) VALUES ('$userName','$email','$password','$countries','$state','$city','$company','$manager1','$manager2')";
    $data = mysql_query ($query)or die(mysql_error());
    if($data)
    {
    echo $state
    //echo "YOUR REGISTRATION IS COMPLETED...";
    }
}

function SignUp()
{
    if(!empty($_POST['userName']))   //checking the 'user' name which is from Sign-Up.html, is it empty or have some text
    {
        if ($_POST['equals_val'] != ($_POST['first_val'] + $_POST['second_val'])) {
            echo "CAPTCHA IS NOT CORRECT";
        } else {
            $query = mysql_query("SELECT * FROM X1 WHERE userName = '$_POST[userName]' AND pass = '$_POST[password]'") or die(mysql_error());

            if(!$row = mysql_fetch_array($query) or die(mysql_error()))
            {
                NewUser();
            }
            else
            {
                echo "SORRY...YOU ARE ALREADY REGISTERED USER...";
            }
        }
    }
}
if(isset($_POST['submit']))
{
    SignUp();
}
if (isset($_POST['type'])) {
    if ($_POST['type'] == "login") 

}
?>
Inline Code Example Here

`

connect.php
`

<?php

$serverName = "localhost";
$userName = "*****";
$password = "**********"; 
$dbName = "******";

// Create connection
$conn = new mysqli($serverName, $userName, $password, $dbName); 

// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

// Prepare and bind

    $userName = $_POST['user'];
    $email = $_POST['email'];
    $password =  $_POST['pass'];
    $countries = $_POST ['countries'];
    $state = $_POST ['state'];
    $city =  $_POST['city'];
    $company = $_POST ['company'];
    $manager1 = $_POST ['manager1'];
    $manager2 = $_POST ['manager2'];

    echo "Thank you for registering - New records created successfully!"; 

?>

`

[Matthew_16]

Update:

Now that I am again looking at other versions of the code, I notice some major issues; there are obviously some elements missing in the code I originally posted here and am using. So, please forgive this oversight. It's embarrasing.

This is another result of dealing with multiple versions; one year ago (when shutting this project down, I thought that I was very thorough about backing up all code for future use) Now, I see I was not thorough enough. I had the feeling this would happen.

I am going to edit my code now and will repost with further questions if necessary .

Thanks!

[Matthew_16]

Update:

Upon adjusting the connect.php file, and testing it, I receive the following response:

Access denied for user '#####'@'localhost' (using password: NO)

The code I added was:

`

 $query = "INSERT INTO ######  (userName,email,password,countries,state,city,company,manager1,manager2) 
    VALUES ('$userName','$email','$password','$countries','$state','$city','$company','$manager1','$manager2')";

    $data = mysql_query ($query)or die(mysql_error());
    if($data)
    {
    echo "YOUR REGISTRATION IS COMPLETED...";
    }Inline Code Example Here

`

I am not sure why I am getting this error message. I have a password set and have never seen this before.

How could adding that snippet of code cause this?

[diafol]

You seem to be mixing mysqli and mysql. Stick to mysqli. They aren.t the same thing.