Error when a function added in an already existing function

Question

I have written a function "attempt_login1()" for login. It works fine.

But when I included function "login_count()" to count total number of login(it's commented out in code) it's giving an error

Parse error: syntax error, unexpected '$connection' (T_VARIABLE) in F:\XAMP1\htdocs\sandbox\test.php on line 57

Am not sure what is that it is causing this errors. Thanks for help.

$username = "user123";
$password = "12345";

attempt_login($username, $password);

    function attempt_login($username, $password){
        $user = find_admin_by_username($username);  //select user by username in database
        $user_set = mysqli_fetch_assoc($user);

            if($user_set){
                $user_password = $user_set["hashed_password"];
                $hashed_password = password_verify($password, $user_password);

                if($hashed_password == true){
            //      login_count($user_set["id"], $user_set["login_count"]);
                    return $user_set;
                }else{
                    return false;
                }
                }else{
                    return false;
                }
            }

    function login_count($user_id, $count){
        $global $connection;

        $count = $count++;
        $query2 = "UPDATE admins SET login_count = $count WHERE id = $user_id";
        $result2 = mysqli_query($connection, $query2);

    }
function find_admin_by_username($username){
    global $connection;

    $query  = "SELECT * ";
    $query .= "FROM admins ";
    $query .= "WHERE username = '{$username}' ";
    $query .= "LIMIT 1";
    $user_set = mysqli_query($connection, $query);
    confirm_query($user_set);

    if($user_set){
        return $user_set;
    }else{
        return NULL;
    }
}

Typo mistake $global $connection should be global $connection , — Drudge Rajen, Feb 29 '16 at 12:02
You should know the difference between data type and variables. — Maha Dev, Feb 29 '16 at 12:06

score 0 · Answer 1 · edited Feb 29 '16 at 15:59

0

You have a spelling mistake:

$global $connection should be global $connection without dollar before global.

edited Feb 29 '16 at 15:59

trincot

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answered Feb 29 '16 at 12:04

Drudge Rajen

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score 0 · Accepted Answer · answered Feb 29 '16 at 12:07

0

Use below code for your function login_count :

function login_count($user_id, $count){
    global $connection;

    $count = $count++;
    $query2 = "UPDATE admins SET login_count = $count WHERE id = $user_id";
    $result2 = mysqli_query($connection, $query2);

}

Hope it will help you :)

answered Feb 29 '16 at 12:07

Milan

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thanks got it @milan – Feb 29 '16 at 12:08

Error when a function added in an already existing function

2 Answers2