（PHP / SQL）不会基于查询参数从数据库中提取

Question

So I'm trying to build a php page to edit records. Right now I'm trying to get it to take the ID in the url and return data based on that. I can echo out the id from the url, that's fine, but the sql query isn't doing anything. I threw the 2 echo's at the bottom to try to get them to display values. The ID displays fine, coming from the URL, but it should display the name corrosponding with that gigid and isn't.

Any ideas?

<?php

 // query db

 $gigid = $_GET['gigid'];
 $result = ORM::for_table('gigs')->where('gigid', $gigid)
 or die(mysql_error()); 
 $row = mysqli_fetch_array($result);

 // check that the 'id' matches up with a row in the databse
 if($row)
 {
 // get data from db
        $gig_name = $row['gig_name'];
        $gig_type = $row['gig_type'];
        $gig_date = $row['gig_date'];
        $gig_customer = $row['gig_customer'];
        $gig_venue = $row['venue_name'];
        $gig_fee = $row['gig_fee'];
        $gig_status = $row['gig_status'];   
 }
mysqli_close($con);
 ?>
 <?php echo $gigid; ?>
 <?php echo $gig_name; ?>

Answer 1

I would just use mysqli_query for any database manipulation.

. . .
$con = mysqli_connect("i.p.addr","username","password","database");
$result = mysqli_query($con, "SELECT * FROM gigs WHERE gigid=$gigid") or die(mysqli_error($con)); 
$row = mysqli_fetch_array($result);
mysqli_close($con);

UPDATE:: changed to mysqli_error()