如何在創建的表的會話消息中回顯變量

Question

我有一個PHP頁面，腳本正在搜索某個特定名稱的表是否存在。如果沒有表格，它會顯示合適的信息。如果存在表格，則會向其添加一些數據並顯示成功會話消息。現在，我想顯示會話消息爲Data added to ABC (name of the chosen table) successfully。相反，我得到的會話消息爲Data added to $table sucessfully。

請幫忙。以下是我的代碼 -

<?php 
require_once("config.php"); 
require_once("config2.php"); 
$keyQ = $db->query("SELECT * FROM Collection where id = 
$Id"); 
$kVal = $kewyQ->fetch(PDO::FETCH_ASSOC); 
$table = trim($kVal['words']); 
try 
{ 
// Select 1 from table_name will return false if the table does not exist. 
$value = $db2->query("select 1 from `$table` LIMIT 1"); 
}catch (PDOException $e) { 
echo "$table does not exist"; 
exit; 
} 
if($value !== FALSE) 
{ 
........ 
........ 
session_start(); 
$_SESSION['success'] = 'Data added to $table successfully.'; 
header('location:list2.php'); 
} 
?> 

Answer 1

使用雙引號而不是單引號來呈現字符串內的變量。

"Data added to $table successfully.";