通过php在页面上显示查询结果

connection to the database

$dbc = mysqli_connect('localhost','root','','') OR die('ERROR :'.mysqli_connect_error());

over here i am using 3 tables,

• farm(farm_id, taluka, village,cropgrown)
• interest(id_interest, talukainvest, villageinvest, cropgrowninvest)
• registration (reg_id,name,....)

i want to output name, taluka, village, crop grown

where taluka=talukainvest, village=villageinvest, cropgrown=cropgrowninvest

and i want this to display in a table form kind of on the php page using browser.

       $z  = "SELECT f.farm_id, 
            r.name,
            f.village,
            i.talukainvest,
            i.cropgrowninvest
        FROM
            farm f
                JOIN interest i
        ON f.village = i.villageinvest
                JOIN registration r
            ON f.farm_id = r.reg_id
        WHERE f.taluka = i.talukainvest AND
              f.cropgrown = i.cropgrowninvest";


    $y = mysqli_query($dbc,$z);
    $x = mysqli_fetch_assoc($y);

    if ($x) {

            echo 'id  :'.$x['farm_id'].'name :'.$x['name'].'village :'.$x['village'].'taluka :'.$x['talukainvest'].'crop grown :'.$x['cropgrowninvest'] ;
        }   

        else{
            echo'queries error'.mysqli_error($dbc);
        }

i tried to test this but its not working.

is there any alternate way than what i have done?if this is not efficient. newbie learning and you guys are helping alot in this.