Why am i getting a fatal error? [duplicate] - php

Question

This question already has answers here:
Reference: What is variable scope, which variables are accessible from where and what are "undefined variable" errors?
(3 answers)
Closed 9 years ago.
Here is the snippet of code that is causing the error.
public function storeUser($email, $password) {
require_once 'include/user.config.inc.php';
$uuid = uniqid('', true);
//echo $uuid;
$hash = $this->hashSSHA($password);
//echo $hash;
$encrypted_password = $hash["encrypted"]; // encrypted password
//echo $encrypted_password;
$salt = $hash["salt"]; // salt
//echo $salt;
$query = "INSERT INTO user_table (unique_id, email_id, encrypted_password, salt, created_at) VALUES ( :uuid, :email, :encrypted_password, :salt, NOW()) ";
$query_params = array(
':uuid' => $uuid,
':email' => $email,
':encrypted_password' => $encrypted_password,
':salt' => $salt
);
try {
//These two statements run the query against the database table.
$stmt = $db -> prepare($query);
$result = $stmt -> execute($query_params);
} catch (PDOException $ex) {
$response["success"] = 0;
$response["message"] = "Database Error! Problem with first query!";
die(json_encode($response));
}
}
I am getting an error:
Notice: Undefined variable: db
Fatal error: Call to a member function prepare() on a non-object
It seems that $query and $query_params is causing the problem but I don't know why. It seems right to me.
Here's the other snippet of code
if ($db->isUserExisted($email)) {
// user is already existed - error response
$response["error"] = 2;
$response["error_msg"] = "User already exist";
echo json_encode($response);
} else {
$db->storeUser($email, $password);
}

Answer 1

Taking a guess here...
require_once only includes the file once, ever*. You have probably required that file before elsewhere, it's not getting loaded again, your $db variable is nowhere to be found.
* In the current script execution.

Answer 2

You never initialize your $db object.
You need something like the following before "$stmt = $db -> prepare($query);"
$db = new mydbclass();