I have to connect to a database twice for a function

Question

At the start of the document, i call include database.php , which contains

$database = new mysqli('localhost', 'root', 'root', 'dbname', 3306);

However, the $database variable doesnt seem to exist when i try to update db info via a function, so i have to call it again

Is there any solution to this?

function code:

<?php
            include 'database.php';        

?>

. . .

<?php 


   function roll() {

$bet= $_POST["bet"];
    $_SESSION['dice1']=rand(1,6);
    $_SESSION['dice2']=rand(1,6);
    $total=$_SESSION['dice1']+$_SESSION['dice2'];



if($total==2) $win=$bet*2;
 else if($total==3 || $total==4) $win=0;
  else if($total==10 || $total==11) $win=$bet;
   else if($total==12) $win=$bet*4;
    else if($total==0) $win=0;
else $win=$bet-(2*$bet);


  //  $database = new mysqli('localhost', 'root', 'root', 'dbname', 3306);
    $database->query("UPDATE users SET balance=balance+'$win' WHERE name='$_SESSION[username]'");

}

I suppose you don't know about variables' scope? Show you function code. — u_mulder, Apr 05 '18 at 06:56
post full code you need to know the var scope as @u_mulder mentioned. — Thamaraiselvam, Apr 05 '18 at 06:57
Your code tabbing is poor, your set of conditionals can be simplified, and pass your connection variable as an argument of `roll()`. — mickmackusa, Apr 05 '18 at 12:01

sunny bhadania · Answer 1 · 2018-04-05T07:21:40.853

 # And this is my database connection file:
    <?php
            $servername = "localhost";
            $username = "root";
            $password = "";
            $dbname = "test";

                // Create connection
                $conn = new mysqli($servername, $username, $password, $dbname);

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

            $conn->close();
        ?>

 # This is my login page:
        <?php 
            include ('database.php');
            include('password_hash_lib/password.php');

            if (isset($_POST['data']))
            {
                $data = $_POST['data'];
                $auth = json_decode($data);
                $user_email = $auth->Email;
                $user_pass = $auth->Password;
                authenticate($user_email, $user_pass);
            }

            function authenticate($Email, $Password)
            {   
                 global $conn;
                 $HashedPassword = password_hash($Password, PASSWORD_DEFAULT);        
                 $sql = "SELECT * FROM app_users WHERE user_email='$Email'"; 
                 $result = $conn->query($sql);
                 $User = $result->fetch_object();

                if ($User->user_email == '')
                {
                    header("Location: login-page.html?msg=failed");
                }

                if (password_verify($Password, $User->user_password_hash))
                {
                    $_SESSION["user_auth_details"] = $User->user_id . "+" . $User->user_email . '+' . $User->user_name . "+" . $User->user_display_image . "+" . $User->user_display_name;
                    header("Location:" . $_SESSION['page_url']);
                }
                 else {
                    header("Location: login-page.html?msg=failed");

                 }
            }

        ?>

What does this answer explain? – u_mulder Apr 05 '18 at 07:08 — u_mulder, Apr 05 '18 at 07:08
Please try it :- global $database; – sunny bhadania Apr 05 '18 at 07:26 — sunny bhadania, Apr 05 '18 at 07:26

I have to connect to a database twice for a function

1 Answers1