PHP登录代码不从数据库读取

Question

我正在尝试为 Android 应用程序创建登录 php，我从教程中修改了此 php，问题是我过去总是得到“成功”= 0“消息”=“并非所有字段都已填充”的响应，所以我添加了一些 print_r 来看看问题出在哪里，现在我只得到这个结果 aimatosnintendo ，其中是用户名和密码的输入，所以它甚至没有到达 ifs，这是我的代码:

<?php
// array for JSON response
$response = array();
define('DB_USER', ""); // db user
define('DB_PASSWORD', ""); // db password (mention your db password here)
define('DB_DATABASE', ""); // database name
define('DB_SERVER', ""); // db server
// array for JSON response

$conn = new mysqli(DB_SERVER, DB_USER, DB_PASSWORD,DB_DATABASE);

// check for post data
print_r ($_POST['username']);
print_r ($_POST['password']);

if(isset($_POST['username'],$_POST['password'])) {
    $username = $_POST["username"];
    $password = $_POST["password"];


$sql = "SELECT *FROM login WHERE username = $username AND password = $password";
$result = $conn->query($sql) or die (mysqli_connect_error());
 print_r ($username);

    if (!empty($result)) {
        // check for empty result

        if (mysqli_num_rows($result) > 0) {

            $result = mysqli_fetch_array($result);

            $loginfo = array();
            $loginfo["name"] = $result["name"];
            $loginfo["username"] = $result["username"];
            $loginfo["password"] = $result["password"];
            $loginfo["phone"] = $result["phone"];
            $loginfo["email"] = $result["email"];
            $loginfo["license"] = $result["license"];
            //$loginfo["expiration"] = $result["expiration"];


            // success
            $response["success"] = 1;

            // user node
            $response["logina"] = array();

            array_push($response["logina"], $loginfo);

            // echoing JSON response
            echo json_encode($response);
        } else {
            // no product found
            $response["success"] = 0;
            $response["message"] = "Wronglogin";

            // echo no users JSON
            echo json_encode($response);
        }
    } else {
        // no product found
        $response["success"] = 0;
        $response["message"] = "Wronglogin";

        // echo no users JSON
        echo json_encode($response);
    }
} else {
    // required field is missing

    $response["success"] = 0;
    $response["message"] = "Not all fields are filled";


    // echoing JSON response
    echo json_encode($response);
}
?>

数据库用户密码服务器和名称都是正确的，为了安全起见，我只是在这篇文章中删除了它们，我有一个在数据库上注册信息的代码，它可以工作，但这个让我陷入困境

我期待 $response["success"] = 1;当用户名和密码与数据库匹配时

Answer 1

尝试

$sql = "SELECT * FROM login WHERE username = '". $username ."' AND password = '".$password."'";