Can't Change mysql_fetch_array to mysqli_fetch_array

Question

I'm working on pagination on a website I'm building and I found a good pagination example here: http://www.awcore.com/dev/1/3/Create-Awesome-PHPMYSQL-Pagination_en . I'd like to us this on my page but am using mysqli instead of mysql and need to convert some of it.

I'm new to MySQL and am trying to figure out the syntax to converting the following code:

function pagination($query, $per_page = 10,$page = 1, $url = '?'){        
    $query = "SELECT COUNT(*) as `num` FROM {$query}";
    $row = mysql_fetch_array(mysql_query($query));
    $total = $row['num'];
    $adjacents = "2"; 

I know it doesn't give you much as to what the entire code is (it's about 100 lines long) but I'm assuming this may be an easy syntax change. I had initially done it likes this:

function pagination($query, $per_page = 10,$page = 1, $url = '?'){        
    $query = "SELECT COUNT(*) as `num` FROM {$query}";
    $row = mysqli_fetch_array(mysqli_query($query));
    $total = $row['num'];
    $adjacents = "2";

I am aware that mysqli_query needs to take two parameters but I was also under the impression that it was the same for mysql_query so I guess I'm just not understanding the code. Sorry if this is a super basic question, I'm just trying to wrap my head around some of these concepts! Thanks for any help.

BTW I did see this question (Converting from mysql to mysqli (mysql_fetch_array)) but it seemed like he was taking several extra steps that may not need to be taken.

EDIT

Here's the error messages I'm getting with the above code FYI:

Warning: mysqli_query() expects at least 2 parameters, 1 given in linkinformation/functions.php on line 9

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given in linkinformation/functions.php on line 9

EDIT

So I added a connection within the function (is that the right approach? I tried connecting outside the function but it was grabbing the info):

function pagination($mysqli, $query, $per_page = 10,$page = 1, $url = '?'){        
    $mysqli = mysqli_connect("localhost","username","password", "db_name");
    $query = "SELECT COUNT(*) as `num` FROM {$query}";
    $row = mysqli_fetch_array(mysqli_query($mysqli, $query));
    $total = $row['num'];
    $adjacents = "2"; 

I get this warning:

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in linkstuff/functions.php on line 10

It's getting a boolean (which I'm assuming is TRUE) whereas it should be getting something else I suppose/the actual query.

Answer 1

//You need to change thses variables

global $conn;  
$conn = new mysqli('localhost','root','',$databasename);

    function pagination($query, $per_page = 10,$page = 1, $url = '?'){        
        $query = "SELECT COUNT(*) as `num` FROM {$query}";
        $result = $GLOBALS['conn']->query($query);
        $row=$result->fetch_array;
        $total = $result->num_rows;
        $adjacents = "2";

Answer 2

Something like this:

//You need to change thses variables    
$conn = new mysqli('localhost','root','',$databasename);

    function pagination( $conn,$query, $per_page = 10,$page = 1, $url = '?'){        
        $query = "SELECT COUNT(*) as `num` FROM {$query}";
        $result = $conn->query($query);
        $row=$result->fetch_array;
        $total = $result->num_rows;
        $adjacents = "2";

Answer 3

You need to change your connection to mysqli as well, and pass the connection object to mysqli_query as the first argument.

function pagination($link, $query, $per_page = 10,$page = 1, $url = '?'){        
    $query = "SELECT COUNT(*) as `num` FROM {$query}";
    $row = mysqli_fetch_array(mysqli_query($link, $query));
    $total = $row['num'];
    $adjacents = "2";