我从下面的处理表单中得到一个错误。我不知道为什么当我点击submit时它没有更新到数据库。它说它已经成功了，但是数据库还没有更新？！
我想这可能是因为“mysqli\u stmt\u execute”，但我不确定。有什么想法吗？/谢谢。

<?php
if(isset($_POST['submit'])){

    include_once 'connection.php';

    $first = mysqli_real_escape_string($conn, $_POST['first']);
    $last = mysqli_real_escape_string($conn, $_POST['last']);
    $uid = mysqli_real_escape_string($conn, $_POST['uid']);
    $pwd = mysqli_real_escape_string($conn, $_POST['pwd']);
    $hashedPwd = md5($pwd); 

     //Error Handling
    //Checking for empty fields  
    if(empty($first) || empty($last) || empty($uid) || empty($pwd)){
        header("Location: ../signup.php?signup=empty");
        exit();
    }else{
        //Checking if the input characters are valid
        if(!preg_match("/^[a-zA-Z]*$/", $first) || !preg_match("/^[a-zA-Z]*$/", $last) ){
            header("Location: ../signup.php?signup=invalid");
            exit();
        }else{
            $sql = "SELECT * FROM users WHERE Staff_ID='$uid'";
            $result = mysqli_query($conn, $sql);
            $resultCheck = mysqli_num_rows($result);
            if($resultCheck > 0){
                header("Location: ../signup.php?signup=usertaken");
                exit();
            }else{
                //Hasing the password surecirty

            }
            }
        }
    //Insert user into database

            $sql = "INSERT INTO users (Firstname, Lastname, Staff_ID, Password) VALUES (?, ?, ?, ?);";

            $stmt = mysqli_stmt_init($conn);

            if (!mysqli_stmt_prepare($stmt, $sql)){
                echo"SQL Error";
                } else{
                    mysqli_stmt_bind_param($stmt, "ssss", $first, $last, $uid, $pwd);
                    mysqli_stmt_execute($stmt);
                    }
                    header("Location: ../signup.php?signup=success");
                    exit();

    }else{
    header("Location: ../signup.php");
    exit();

}
?>