i have this connection

    $db_host = 'localhost';
    $db_user = 'root';
    $db_pass = '';
    $db_name = 'databasename';      

    $conn = mysql_connect($db_host,$db_user,$db_pass) or die(mysql_error());
mysql_query("SET NAMES 'utf8'");
mysql_query('SET CHARACTER SET utf8');  

but my old connection dosen't work with this code

//connect to database
$link = mysqli_connect('localhost','root','');

//get all rows
$query = mysqli_query($link,'SELECT * FROM categories');
while ( $row = mysqli_fetch_assoc($query) )
    $menu_array[$row['catid']] = array('catname' => $row['catname'],'parentid' => $row['parentid']);
    $menu_array[$row['catid']] = array('catname' => $row['catname'],'parentid' => $row['parentid'],'catid'=>$row['catid']);

//recursive function that prints categories as a nested html unorderd list
function generate_menu($parent)
    $has_childs = false;
    //this prevents printing 'ul' if we don't have subcategories for this category

    global $menu_array;
    //use global array variable instead of a local variable to lower stack memory requierment

    foreach($menu_array as $key => $value)
        if ($value['parentid'] == $parent) 
            //if this is the first child print '<ul>'            
            if ($has_childs === false)
                //don't print '<ul>' multiple times              
                $has_childs = true;
                echo '<ul>';
            echo '<li><a href="category.php?catid='. $value['catid'] . '">' . $value['catname'] . '</a>';
            //call function again to generate nested list for subcategories belonging to this category
            echo '</li>';
    if ($has_childs === true) echo '</ul>';

I think My erro located between this code .. Finally i need just replace mysqli_connect to mysql_connect Without I

while ( $row = mysqli_fetch_assoc($query) )
    $menu_array[$row['catid']] = array('catname' => $row['catname'],'parentid' => $row['parentid']);

THERE is differenct between parameter passing in mysql and mysqli. Link is passed later in mysql and passed as first parameter in mysqli.
so refer manual of mysql and replace parameter position where applicable

thanks for you but i'm try to do it but i do'nt understand :S any exampls ??!!

yes, he's saying you need to pass the database name in mysqli

like this:
$db = new mysqli($host, $user, $pass, $dbname) or die($error);

LOL i had to lookup that acronym, 1st encounter.

Funny and true.