如果在SQLite中使用外键，则创建表失败(Create table fails if foreign keys are used in SQLite)

 我真的不知道为什么这个PHP代码不创建三个表。 第一和第二都很好。 但第三个失败了。 第三个与第二个非常相似，但在外键定义之后有另一个字段。 我在这做错了什么？ 
 
$db = new \PDO("sqlite:d:/temp/test_db.sqlite");
$db->exec("PRAGMA foreign_keys = 'ON'");


$statement = $db->query("CREATE TABLE  IF NOT EXISTS customers(
  id TEXT PRIMARY KEY NOT NULL , 
  name TEXT  
)");

$statement->execute();

$statement = $db->query("CREATE TABLE  IF NOT EXISTS appointments (
  id TEXT PRIMARY KEY NOT NULL , 
  customer TEXT   , 
  FOREIGN KEY (customer) REFERENCES customers(id)  
)");

$statement->execute();


$statement = $db->query("CREATE TABLE  IF NOT EXISTS appointment (
  id TEXT PRIMARY KEY NOT NULL ,  
  customer TEXT   , 
  FOREIGN KEY (customer) REFERENCES customers(id),
  nextfield TEXT
)");

$statement->execute();

I seriously have no idea why this php code is not creating three tables. The first and second are fine. But the third fails. The third is very similar to the second but has another field after the foreign key definition. What am I doing wrong here? 
 
$db = new \PDO("sqlite:d:/temp/test_db.sqlite");
$db->exec("PRAGMA foreign_keys = 'ON'");


$statement = $db->query("CREATE TABLE  IF NOT EXISTS customers(
  id TEXT PRIMARY KEY NOT NULL , 
  name TEXT  
)");

$statement->execute();

$statement = $db->query("CREATE TABLE  IF NOT EXISTS appointments (
  id TEXT PRIMARY KEY NOT NULL , 
  customer TEXT   , 
  FOREIGN KEY (customer) REFERENCES customers(id)  
)");

$statement->execute();


$statement = $db->query("CREATE TABLE  IF NOT EXISTS appointment (
  id TEXT PRIMARY KEY NOT NULL ,  
  customer TEXT   , 
  FOREIGN KEY (customer) REFERENCES customers(id),
  nextfield TEXT
)");

$statement->execute();

原文：https://stackoverflow.com/questions/48019564