php - 获取 SQL 语法错误

Question

我试图在我的 php 代码中从我的应用程序中获取这两个值。但在这样做之前，我试图通过 URL 检查。但我的问题是，如果给出值手册，我会得到正确的输出，但是当我通过传递值来检查它时，我会收到语法错误。谁能帮我解决这个问题。

<?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);
} 



?>

Answer 1

I am trying to check through URL

据此，我假设您的意思是您正在尝试访问 url:

http://localhost/yoursite/yourpage?Qno=5&day=thing

在这种情况下，这些变量将作为 $_GET['Qno'] 和 $_GET['day'] 访问。

您可以使用 $_REQUEST['Qno'] 和 $_REQUEST['day'] 两种方式接收变量。当然，您的应用程序有很多安全漏洞，我什至都不会去碰。