Question

我真的不知道为什么这个PHP代码没有创建三个表。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();

Answer 1

根据SQLite grammar规范，nextfield TEXT应与其他列定义放在一起。 FOREIGN KEY与table-constraint部分相关，应在列定义后定义：

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

Answer 2

FOREIGN KEY必须在列之后进行编码，或者作为列定义的一部分进行编码而不使用FOREIGN KEY（列）作为隐含的列，因此替换

"CREATE TABLE  IF NOT EXISTS appointment (
  id TEXT PRIMARY KEY NOT NULL ,  
  customer TEXT   , 
  FOREIGN KEY (customer) REFERENCES customers(id),
  nextfield TEXT
)"

with（table constraint）： -

"CREATE TABLE  IF NOT EXISTS appointment (
  id TEXT PRIMARY KEY NOT NULL ,  
  customer TEXT   , 
  nextfield TEXT ,
  FOREIGN KEY (customer) REFERENCES customers(id))"

或（列约束）： -

"CREATE TABLE  IF NOT EXISTS appointment (
  id TEXT PRIMARY KEY NOT NULL ,  
  customer TEXT  REFERENCES customers(id) , 
  nextfield TEXT)"

如果在SQLite中使用外键，则创建表失败

2 个答案: