PHP แก้โค้ดจาก php 5 มาเป็น 7 ไม่ได้ครับมันขึ้น Notice: Undefined variable

<?php

$Action = $_GET["Action"];
$QuestionID = $_GET["QuestionID"];
$txtDetails = $_POST["txtDetails"];
$txtName = $_POST["txtName"];

$dbConn = mysqli_connect ($dbHost, $dbUser, $dbPass , $dbName) or die ('MySQL connect failed. ' . mysqli_error());

if($Action == "Save")
{
	//*** Insert Reply ***//
	$strSQL = "INSERT INTO tbl_webboard_reply ";
	$strSQL .="(QuestionID,CreateDate,Details,Name) ";
	$strSQL .="VALUES ";
	$strSQL .="( $QuestionID ,'".date("Y-m-d H:i:s")."', $txtDetails , $txtName)";
	$objQuery = mysqli_query($dbConn, $strSQL) or die(mysqli_error($dbConn));
	
	//*** Update Reply ***//
	$strSQL = "UPDATE tbl_webboard_question ";
	$strSQL .="SET Reply = Reply + 1 WHERE QuestionID = $QuestionID ";
	$objQuery = mysqli_query($dbConn, $strSQL) or die(mysqli_error($dbConn));	
}

//*** Select Question ***//
$strSQL = "SELECT * FROM tbl_webboard_question WHERE QuestionID =  $QuestionID ";
$objQuery = mysqli_query($dbConn, $strSQL) or die(mysqli_error($dbConn));
$objResult = mysqli_fetch_array($objQuery,MYSQLI_ASSOC);



//*** Update View ***//
$strSQL = "UPDATE tbl_webboard_question ";
$strSQL .="SET View = View + 1 WHERE QuestionID = $QuestionID ";
$objQuery = mysqli_query($dbConn, $strSQL) or die(mysqli_error($dbConn));	
?>
<table width="738" border="1" cellpadding="1" cellspacing="1">
  <tr>
    <td colspan="2"><center><h1><?php echo $objResult['Question'];?></h1></center></td>
  </tr>
  <tr>
    <td height="53" colspan="2"><?php echo nl2br($objResult['Details']);?></td>
  </tr>
  <tr>
    <td width="397">Name : <?=$objResult["Name"];?> Create Date : <?=$objResult["CreateDate"];?></td>
    <td width="253">View : <?=$objResult["View"];?> Reply : <?=$objResult["Reply"];?></td>
  </tr>
</table>