这个 mysqli 查询有什么问题

我试图弄清楚下面的mysqli代码出了什么问题。数据肯定在表中。我有两个问题：

1. 为什么这没有给我任何结果？

2. 我可以将回显<table></table>部分放在单独的 php 文件中吗，如果是，我将如何调用$name,$partner,$cell等？

     <?php
     $mysqli = new mysqli($host, $uname, $pword, $db);
   if ($mysqli->connect_errno)
   {
   echo "Connection failed: %s'n", $mysqli->connect_error;
   exit();
   }
     $query = ("SELECT * FROM contacts WHERE contacttype IN ('Buyer','Seller','Buyer / Seller','Investor') AND leadstatus = 'New' ORDER BY date DESC");
     $result = $mysqli->query($query);
     while ($row = $result->fetch_array()) {
     $space = (!empty($row['firstname']) && !empty($row['lastname'])) ? ' ' : '';        
     $name = $row['firstname'].$space.$row['lastname'];       
     $partner  = $row['spousefirst'];   
     $cell = (!empty($row['phonecell'])) ? " {$row['phonecell']} (cell)" : ''; 
     $email = (!empty($row['email'])) ? " {$row['email']} (email)" : '';
     $ID = $row['ID'];
     echo'<table>
     <tbody>
     <tr>
       <td><input type="checkbox" name="" id="" value="<? echo $ID; ?>"></td>
       <td><a href="/backend/leads/view/?ID=<? echo $ID; ?> "><strong><? echo $name; ?></strong></a></td>
       <td><a href="/backend/leads/view/?ID=<? echo $ID; ?> "><? echo $partner; ?></a></td>
       <td><? echo $phonecell; ?></td>
       <td><a href="mailto:<? echo $email; ?>"><? echo $email; ?></a></td>
       <td><? echo date("M jS, g:i A", strtotime($date)); ?></td>
       <td><? echo $contacttype; ?></td>
       <td><? echo $agentassigned; ?></td>
       <td><? echo $leadstatus; ?></td>
       <td><a href="/backend/contacts/notes.php?ID=<? echo $ID; ?>">View </a>+ </td>
       <td><a href="/backend/contacts/todo.php?ID=<? echo $ID; ?>">View </a>+ </td>
       <td><a href="/backend/contacts/deletesuccess.php?ID=<? echo $ID; ?>">D</a></td>
       </tr>
       </tbody>
   </table>';
        }
        ?>
   

编辑：

它显示了<table>，但只是没有给我$name的价值，$email......等。