Problem z utworzeniem bazy Opcje

[dejmen]

Otóż chcę utworzyć bazę z poziomu php oto skrypt

[PHP] pobierz, plaintext 
<?php
$connect = mysql_connect("localhost", "root", "mysql");
 
 
$create = mysql_query("CREATE DATABASE IF NOT EXISTS moviesite")
or die(mysql_error());
 
mysql_select_db("moviesite");
 
$movie = "CREATE TABLE movie (
movie_id int(11) NOT NULL auto_increment,
movie_name varchar(255) NOTNULL,
movie_type tinyint(2) NOT NULL default 0,
movie_year int(4) NOT NULL default 0,
movie_leadactor int(11) NOT NULL default 0,
movie_director int(11) NOT NULL default 0,
PRIMARY KEY (movie_id),
KEY movie_type (movie_type,movie_year)
)";
 
$results = mysql_query($movie)
or die (mysql_error());
 
$movietype = "CREATE TABLE movietype (
movietype_id int(11) NOT NUL auto_increment,
movietype_label varchar(100) NOT NULL,
PRIMARY KEY (movietype_id)
)";
 
$results = mysql_query($movietype)
or die (mysql_error());
 
$people = "CREATE TABLE people (
people_id int(11) NOT NULL auto_increment,
people_fullname varchar (255) NOT NULL,
people_isactor tinyint(1) NOT NULL default 0,
people_isdirector tinyint(1) NOT NULL default 0,
PRIMARY KEY (people_id)
)";
 
$results = mysql_query($people)
or die(mysql_error());
 
echo "Poprawnie utworzono baze filmów";
?>
[PHP] pobierz, plaintext 

gdy próbuje otworzyć plik wyskakuj taki błąd :
"You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'NOTNULL, movie_type tinyint(2) NOT NULL default 0, movie_year int(4) NOT NULL ' at line 3"
nie umiem znaleść źródła błędu bardzo proszę o pomoc

[blooregard]

movie_name varchar(255) NOTNULL,

NOT NULL