Question

我有一个与登录页面一起使用的if语句。

我一直在尝试将其转换为mysqli预备语句。以前只是正常的mysqli。

问题在于，无论我输入什么ID，它都会一直说无效ID！而不是切换到正确的页面。

if (isset($_POST['login']))
{
require "connect.php";

session_start();

if (count($_POST) > 0)
    {
    if ($stmt = mysqli_prepare($conn, "SELECT id, Login_ID, Name, User_Role_ID FROM user WHERE Login_ID = ?"));

    $lid = $_POST["id"];

    $stmt->bind_param("i", $lid);

    $stmt->execute();

    $stmt->bind_result($id, $Login_ID, $Name, $User_Role_ID);

        $_SESSION["Student_DB_ID"] = $id;
        $_SESSION["Login_ID"] = $Login_ID;
        $_SESSION["Name"] = $Name;
        $_SESSION["User_Role_ID"] = $User_Role_ID;

        switch ($User_Role_ID)
            {
        case "2":
            header("Location: ../views/student/");
            break; //Student
        case "1":
            header("Location: ../views/admin/");
            break; //Admin
        default:
                echo "Invalid ID!"; 
            } 

/* close statement */
$stmt->close();
$conn->close();

    }
}

Answer 1

您需要阅读mysqli_stmt_fetch（）函数（http://php.net/manual/en/mysqli-stmt.fetch.php）的文档。该函数仅返回布尔值，具体取决于查询是否成功完成。您需要使用bind_result（）来返回值。

编辑：

根据下面的评论，此代码应该有效，您还删除了仍然需要的fetch（）调用：

if (isset($_POST['login'])) {
    require "connect.php";

    session_start();

    if (count($_POST) > 0) {
        if ($stmt = mysqli_prepare($conn, "SELECT id, Login_ID, Name, User_Role_ID FROM user WHERE Login_ID = ?")) {
            $lid = $_POST["id"];

            $stmt->bind_param("i", $lid);
            $stmt->execute();
            $stmt->bind_result($id, $Login_ID, $Name, $User_Role_ID);
            $stmt->fetch();

            $_SESSION["Student_DB_ID"] = $id;
            $_SESSION["Login_ID"] = $Login_ID;
            $_SESSION["Name"] = $Name;
            $_SESSION["User_Role_ID"] = $User_Role_ID;

            switch ($User_Role_ID) {
                case "2":
                    header("Location: ../views/student/");
                    break; //Student
                case "1":
                    header("Location: ../views/admin/");
                    break; //Admin
                default:
                    echo "Invalid ID!"; 
            }

            /* close statement */
            $stmt->close();
            $conn->close();
        }
    }
}

Mysqli准备好的声明没有意外

1 个答案: