Notice: Undefined variable: rezistent. Where rezistent is refering to database

Question

I am getting the mentioned error. I have the database connection in conn.php which I am including in fn.php. However, on a particular line it is giving the error

Notice: Undefined variable: rezistent in fn.php on line 16

Line 16 is:

    $a = mysqli_query($rezistent, "SELECT is_verified FROM users WHERE veri_key = '$key'") or die(mysqli_error());

Here's my conn.php

    <?php 
    $user = "mmoin";
    $pass = "pass";
    $host = "localhost";
    $dbname = "rezistent";

    $rezistent = mysqli_connect($host, $user, $pass, $dbname) or die("cannot connect with database");
    ?>

and here's the fn.php

    <?php
    include "conn.php";

    function hashit($v){
        $hash = md5($v);
        $hash .= rand(11,99);
        return $hash;
    }

    function user_verification($k){
    $account_type = substr($k, 0, 1);
    $key = substr($k, 1);
    $msg_to_display = "";

    if($account_type == "h" || $account_type == "t"){
        $a = mysqli_query($rezistent, "SELECT is_verified FROM users WHERE veri_key = '$key'") or die(mysqli_error());

        $rows = mysqli_num_rows($a);
        if($rows > 0){
            mysqli_query($rezistent, "UPDATE users SET is_verified = '1' WHERE veri_key='$key'");
            $msg_to_display = "User successfully verified. Please use the login link to login with your credentials.";
        }
    }
    else{
        $msg_to_display = "There seems to be a problem with the verification key. Please try again from the link provided in the email.";
    }

    return $msg_to_display;
    }
    ?>

What can be the problem? I have tried even connecting with database immediately before the function but it still gives the same notice message.

`mysqli_query()` expects the query as the first parameter and the resource as the second, so your parameters are reversed. But I don't think that's the problem. Have you tried to `echo` $rezistent` to see if it returns a resource handle? — Nick Coons, Aug 04 '13 at 07:19
To clarify on @Horen's comment, the value `$rezistent` isn't within the scope of your function. You'll either need to include it as he suggests, pass it as a parameter to the function, or pull it in as a global. — Nick Coons, Aug 04 '13 at 07:20
okay, great. But isn't this include something global in php? — Muhammad Talha Moin, Aug 04 '13 at 07:20

score 1 · Answer 1 · answered Aug 04 '13 at 07:19

that is a notice, not an error.
that variable really is not defined (you can rely on php that it is not if php says so...).
you declare that variable on top level, but try to use it inside a function. That won't work. You'd have to access the variable as a global variable. Currently it is interpreted as a variable local to the function, where it does not exist. Better even: hand over the variable as a parameter to the function.

score 0 · Answer 2 · answered Aug 04 '13 at 07:19

This is because $rezistent not defined inside the scope of the function user_verification, this is not mean that database connection refused.

Solution:

Send $rezistent as second parameter to the function.

function user_verification($k, $rezistent){

Notice: Undefined variable: rezistent. Where rezistent is refering to database

2 Answers2