用于更新数据的按钮

我有一个按钮列表，每个按钮传递不同的值。代码应该将该值存储为变量或会话，然后将其传递给更新该值的表行的函数。我试着把这个值存储为一个变量，然后传递给它，但这不起作用，所以我想我会把它变成一个会话。那也没用，我不知道该怎么办。

下面是一个按钮示例:

<tr>
  <td data-title="Name"><a href="#"><img src="banner.jpg" width="400"></a></td>
  <td data-title="Link">
    <form name="first" action="pending.php" method="POST"><input type="hidden" name="xz" value="one"><a class="mbutton" onclick="document.forms['first'].submit();">Gedaan</a></form>
  </td>
  <td data-title="Status"><p class="ptable"><?= $fgmembersite->One(); ?></p></td>
</tr>

表单指向:

<?PHP
include("./include/membersite_config.php");
$_SESSION['postdata'] = $_POST;
$bname = $_SESSION['postdata'];
if($fgmembersite->Pending($bname))
{
    $fgmembersite->RedirectToURL("index.php");
}

?>

所以这应该把$bname传递给Pending()函数。

这里是pending:

function Pending($bname)
    {
        if(!$this->DBLogin())
        {
            $this->HandleError("Database login failed!");
            return false;
        }   
        $result = mysql_query("Select * from $this->tablename where confirmcode='y'",$this->connection);   
        if(!$result || mysql_num_rows($result) <= 0)
        {
            $this->HandleError("Error");
            return false;
        }
        $row = mysql_fetch_assoc($result);
        $ux = $row['username'];
        $ustat = $row['one'];
        $bname = $_SESSION['postdata']['xz'];
        $qry = "Update $this->tablename Set $bname='In behandeling' Where  username='$ux'";
        if(!mysql_query( $qry ,$this->connection))
        {
            $this->HandleDBError("Error inserting data to the table'nquery:$qry");
            return false;
        }      
        return true;
    }