获取 SQL 语法错误

我正在尝试从我的php代码中的应用程序中获取这两个值。 但在这样做之前，我正在尝试通过 URL 进行检查。但我的问题是，如果给值手册，我会得到正确的输出，但是当我通过传递值来检查它时，我得到了语法错误。任何人都可以帮助我解决这个问题。

<?php
$hostname_localhost ="localhost";
$database_localhost ="mobiledb";
$username_localhost ="root";
$password_localhost ="";
$localhost = mysql_connect($hostname_localhost,$username_localhost,$password_localhost)
or
trigger_error(mysql_error(),E_USER_ERROR);
 $response = array();
mysql_select_db($database_localhost, $localhost);
$day = $_POST['day'];
$Q = $_POST['Qno'];

    // get a product from products table
$result = mysql_query("SELECT $Q FROM `Questions` WHERE `day`='$day'") or die(mysql_error()); 
    //echo $result;
if (mysql_num_rows($result) > 0) {
    // looping through all results
    // products node
    $response["question"] = array();
    while ($row = mysql_fetch_array($result)) {
        // temp user array
        $product = array();

        $product["question".$i] = $row["$Q"];
     $i = $i + 1;
     // push single product into final response array
        array_push($response["question"], $product);

    }

    // success
    $response["success"] = 1;
    // echoing JSON response
    echo json_encode($response);
} else {
    // no products found
    $response["success"] = 0;
    $response["message"] = "No users found";
    // echo no users JSON
    echo json_encode($response);
} 

?>