mysqli_real_escape_string() expects parameter 1 to be mysqli; mysql_result() expects parameter 1 to be resource

Question

I'm getting these errors:

Notice: Undefined variable: link in login.php on line 11

Warning: mysqli_real_escape_string() expects parameter 1 to be mysqli, null given in login.php on line 11

Notice: Undefined variable: link in login.php on line 18

Warning: mysqli_query() expects parameter 1 to be mysqli, null given in login.php on line 18

Warning: mysql_result() expects parameter 1 to be resource, null given in login.php on line 19

Here's the code:

<?php

session_start();


$link = mysqli_connect('localhost','root','') or die(); 
mysqli_select_db($link,'lr') or die();  


function sanitize($data) {
    return mysqli_real_escape_string($link, $data);     <<line 11
}



function user_exists($username) {
    $username = sanitize($username);    
    $query = mysqli_query($link, "SELECT COUNT('user_id') FROM 'users' WHERE 'username' = '$username'");        <<line 18
    return (mysql_result($query, 0) ==1) ? true : false;        <<line19

}


if (empty($_POST) === false) {
    $username = $_POST['username'];
    $password = $_POST['password'];

    if (empty($username) === true || empty($password) === true) {
        $errors[] = 'You need to enter a username and password';    

    } else if (user_exists($username) === false) {  
        $errors[] = 'User not found';
        }


}

?>

What's the problem?

possible duplicate of [Reference - What does this error mean in PHP?](http://stackoverflow.com/questions/12769982/reference-what-does-this-error-mean-in-php) — deceze, Sep 19 '13 at 06:57
[Reference: What is variable scope, which variables are accessible from where and what are “undefined variable” errors?](http://stackoverflow.com/questions/16959576/reference-what-is-variable-scope-which-variables-are-accessible-from-where-and) — deceze, Sep 19 '13 at 06:57

score 0 · Accepted Answer · answered Sep 19 '13 at 06:47

0

You need to use $_GLOBALS['link'] inside function.

Like this:

    function sanitize($data) {
    return mysqli_real_escape_string($_GLOBALS['$link'], $data);     <<line 11
}

Same thing inside user_exist function.

answered Sep 19 '13 at 06:47

user2770029

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score 0 · Answer 2 · answered Jan 22 '15 at 21:47

Global may have worked to fix the problem, but it's better to pass the variable into the function.

function sanitize($data, $A) {
    return mysqli_real_escape_string($A, $data);     <<line 11
}
function user_exists($username, $A) {
    $username = sanitize($username);    
    $query = mysqli_query($A, "SELECT COUNT('user_id') FROM 'users' WHERE 'usernam' = '$username'");        <<line 18
    return (mysql_result($query, 0) ==1) ? true : false;        <<line19

}

if (empty($_POST) === false) {
    $username = $_POST['username'];
    $password = $_POST['password'];

    if (empty($username, $link) === true || empty($password) === true) {
        $errors[] = 'You need to enter a username and password';    

    } else if (user_exists($username, $link) === false) {  
        $errors[] = 'User not found';
        }


}

mysqli_real_escape_string() expects parameter 1 to be mysqli; mysql_result() expects parameter 1 to be resource

2 Answers2