我的PHP代码似乎在第21行有一个错误检查下面

Question

这里检查出似乎显示错误的代码“//检查输入字符是否有效错误是{括号”我试图通过删除括号或添加另一个来找到错误但它赢了'帮忙。我是php的新手，所以如果可以，请帮帮我;

    <?php
if (isset($_POST['submit'])) {
include_once 'dbh.inc.php';

$first = mysqli_real_escape_string($conn, $_POST['first'] );
$last = mysqli_real_escape_string($conn, $_POST['last'] );
$email = mysqli_real_escape_string($conn, $_POST['email'] );
$uid = mysqli_real_escape_string($conn, $_POST['uid'] );
$pwd = mysqli_real_escape_string($conn, $_POST['pwd'] );
//Error Handlers
//Check for empty fields
if (empty($first) || empty($last) || empty($email) || empty($uid) || 
empty($pwd) ){  
header("Location: ../signup.php?signup=empty");
echo "*Please fill in all fields";
exit();
} else{
//Check if input characters are valid
if (!preg_match("/^[a-zA-Z]*$/", $first) || (!preg_match("/^[a-zA-Z]*$/", 
$last)) {
header("Location: ../signup.php?signup=invalid_input_char");
echo "*Please use Uppercase or Lowercase alphabets";
exit();
}else{
//Check if email is valid
if (!filter_var($email, FILTER_VALIDATE_EMAIL)) {
header("Location: ../signup.php?signup=invalid_email");
echo "*Invalid E-mail address";
exit();
}else{
$sql = "SELECT * FROM users WHERE user_uid= '$uid'";
$result = mysqli_query($conn, $sql);
$resultCheck = mysqli_num_rows($result);

if ($resultCheck > 0) {
header("Location: ../signup.php?signup=useridtaken");
echo "*User Id already taken";
exit();
}
}else{
//Hashed Password
$hashedPwd = password_hash($pwd, PASSWORD_DEFAULT);
//Insert the user into the database
$sql = "INSERT INTO users (user_first, user_last, user_uid, user_pwd) VALUES 
('$first', '$last', '$email', '$uid', '$hashedPwd', );";
mysqli_query($conn, $sql);
header("Location: ../signup.php?signup=success");
echo "Congrats your Account has been Created";
exit();

}
}
}

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