包含数据库值和用户输入的表

所以我已经断断续续地与这个问题斗争了很长一段时间，就像在一个陌生的城市里开车迷路一样，我终于崩溃了！我正在使用数据库中的值开发表，但也需要一个处理用户输入的列。我已经能够显示该表，但我的输入没有更新必要的数据库元素。以下代码：

<?php
include("pogsatbetbuddy.inc.php");        
$cxn=mysqli_connect($host,$username,$password,$db_name)
    or die("Did Not Connect");
$query="SELECT * FROM $tbl2_name ORDER BY $tbl2_name.$col_name ASC";
$result=mysqli_query($cxn,$query)
    or die("Query Not Working");      
echo"<table border='1'
          <form name='payments' action='' method='POST'>
           <tr>
             <td class='update' colspan='5'>
              <button data-theme='b' id='submit' type='submit'>Update</button>
             </td>
            </tr>
    <tr>
         <th class='profile'>Last Name</th>
         <th class='profile'>First Name</th>
     <th class='profile'>Saturday Payment Owing</th>
             <th class='profile'>Enter Payment</th>
             <th class='profile'>Saturday Balance</th>                   
    </tr>";
     while ($row=mysqli_fetch_assoc($result))
      {
   extract ($row);
   echo"<tr>
         <td class='profile'>$lastname</td>
         <td class='profile'>$firstname</td>
         <td class='profile'>$owingsat</td>
             <td class='profile'><input type='number' name='paidsat' value=''/></td>
             <td class='profile'>$owingsat-$paidsat</td>                 
        </tr>";
       }
     echo "</form>";
 echo "</table>";       

这将以我想要的方式显示表格。在处理了以下代码的结果后，我似乎返回了一个null值，所以我认为我对表单操作或提交更新按钮有问题，但经过大量实验和搜索后找不到解决方案。以下代码余额：

if(isset($_POST['paidsat']))
 {
  $paidsat = $_POST['paidsat'];
    if(($paidsat) != null)
     {
       $stmt = $cxn->prepare("UPDATE $tbl2_name SET paidsat = ? WHERE firstname=? and lastname=?");
       $stmt->bind_param('sss', $paidsat, $firstname, $lastname);
       $status = $stmt->execute();       
        if($status === true)  //To check if the execute was successful
        {
         echo("<p class='click'>You have successfully added the payment for $firstname $lastname'n<br /></p>");
        }
      }
   else echo"Not Successful";
  }
 else echo "<p class='click'>Make your changes as required</p>";
 mysqli_close($cxn);

在第二个if语句时，一切都戛然而止。。。。。或者我应该说，虽然东西看起来很漂亮，但它们不起作用！提前感谢，感谢您的帮助！