Php 如何在名称参数（？）中添加特定值

是否可以在名称参数中添加特定值？我给你举个例子

我想在数据库中添加一个特定值的用户，而不需要任何post方法或其他东西。我只想在数据库中添加一个特定的值

$sql = "INSERT INTO accounts (username, email, user_type, password) VALUES(?, users, ?, ?);";

或者还有其他的工作吗

这是全部代码

    // register user if there are no errors in the form
    if (count($errors) == 0) {


    if (isset($_POST['user_type'])) {
        $user_type = e($_POST['user_type']);
        $sql = "INSERT INTO accounts (username, email, user_type, password) VALUES(?, ?, ?, ?);";
        // where going to initialize the new prepared statement using the connection to database
        $stmt = mysqli_stmt_init($db);
        //check of initialize
        if (!mysqli_stmt_prepare($stmt, $sql)) {
            header("location: index.php?error=stmtfailed");
            exit(0);
        }
        $hashedpwd = password_hash($password, PASSWORD_DEFAULT);

        mysqli_stmt_bind_param($stmt, "ssss", $username, $email, $user_type, $hashedpwd);
        mysqli_stmt_execute($stmt);
        mysqli_stmt_close($stmt);


        $_SESSION['add']  = "Added successfully";
        header('location: users.php');
        exit(0);
    }else{
        $sql = "INSERT INTO accounts (username, email, user_type, password) VALUES(?, ?, ?, ?);";
        // where going to initialize the new prepared statement using the connection to database
        $stmt = mysqli_stmt_init($db);
        //check of initialize
        if (!mysqli_stmt_prepare($stmt, $sql)) {
            header("location: index.php?error=stmtfailed");
            exit(0);
        }
        $hashedpwd = password_hash($password, PASSWORD_DEFAULT);

        mysqli_stmt_bind_param($stmt, "ssss", $username, $email, 'user', $hashedpwd);
        mysqli_stmt_execute($stmt);
        mysqli_stmt_close($stmt);

        // get id of the created user
        $logged_in_user_id = mysqli_insert_id($db);
        $_SESSION['user'] = getUserById($logged_in_user_id); // put logged in user in session
        $_SESSION['add']  = "You are now logged in and thank you!";

        header('location: index.php');
        exit(0);
    }
    }

    }

我还尝试在我的注册表单中添加一个隐藏的输入，但它只将用户重定向为管理员，我不希望发生这种情况

这部分代码是针对管理员的。当用户类型为isset时，它是管理员，因为管理员只能在用户或管理员时分配。另一个是给用户的

       if (isset($_POST['user_type'])) {
        $user_type = e($_POST['user_type']);
        $sql = "INSERT INTO accounts (username, email, user_type, password) VALUES(?, ?, ?, ?);";
        // where going to initialize the new prepared statement using the connection to database
        $stmt = mysqli_stmt_init($db);
        //check of initialize
        if (!mysqli_stmt_prepare($stmt, $sql)) {
            header("location: index.php?error=stmtfailed");
            exit(0);
        }
        $hashedpwd = password_hash($password, PASSWORD_DEFAULT);

        mysqli_stmt_bind_param($stmt, "ssss", $username, $email, $user_type, $hashedpwd);
        mysqli_stmt_execute($stmt);
        mysqli_stmt_close($stmt);


        $_SESSION['add']  = "Added successfully";
        header('location: users.php');
        exit(0);
    }

---

事实上，达曼回答了我的问题，我不知道如何报答。我只需要添加$user\u type=“user”

什么是函数

e（）

？您可能应该删除它，因为它会损坏您的数据。如果该值是固定的，您可以在代码中对其进行硬编码，如

$user\u type='admin'你的问题是什么？它是转义字符串的函数吗？为什么要转义字符串？你逃避了什么？也谢谢你给我的参考开始我也在寻找那个！