Question

我正在尝试使用循环依赖来复制Peewee示例应用程序，如the docs中所述，尽管创建者明确指出这通常是个坏主意。这是主要从文档复制的代码：

from peewee import *

db = SqliteDatabase(None)

class BaseModel(Model):

class Meta:
    database = db

class User(BaseModel):
    username = CharField()
    # Tweet has not been defined yet so use the deferred reference.
    favorite_tweet = DeferredForeignKey('Tweet', null=True)

class Tweet(BaseModel):
    message = TextField()
    user = ForeignKeyField(User, backref='tweets')


db.init('twitter.db')
db.create_tables([User, Tweet])
User._schema.create_foreign_key(User.favorite_tweet) #Error
db.close()

我在用#Error注释的行中得到了一个异常。如文档中所述，此行是必需的：

当您调用create_table时，我们将再次遇到相同的问题。对于   因此，peewee不会自动创建外键   任何延迟的外键的约束。

要创建表和外键约束，可以使用   SchemaManager.create_foreign_key（）方法创建约束   在创建表之后。

这是我使用Python 3.5.2的例外情况：

Traceback (most recent call last):
  File "/usr/local/lib/python3.5/dist-packages/peewee.py", line 2653, in execute_sql
    cursor.execute(sql, params or ())
sqlite3.OperationalError: near "CONSTRAINT": syntax error

During handling of the above exception, another exception occurred:

Traceback (most recent call last):
  File "test3.py", line 23, in <module>
    User._schema.create_foreign_key(User.favorite_tweet)
  File "/usr/local/lib/python3.5/dist-packages/peewee.py", line 4930, in create_foreign_key
    self.database.execute(self._create_foreign_key(field))
  File "/usr/local/lib/python3.5/dist-packages/peewee.py", line 2666, in execute
    return self.execute_sql(sql, params, commit=commit)
  File "/usr/local/lib/python3.5/dist-packages/peewee.py", line 2660, in execute_sql
    self.commit()
  File "/usr/local/lib/python3.5/dist-packages/peewee.py", line 2451, in __exit__    reraise(new_type, new_type(*exc_args), traceback)
  File "/usr/local/lib/python3.5/dist-packages/peewee.py", line 178, in reraise
    raise value.with_traceback(tb)
  File "/usr/local/lib/python3.5/dist-packages/peewee.py", line 2653, in execute_sql
    cursor.execute(sql, params or ())
peewee.OperationalError: near "CONSTRAINT": syntax error

Answer 1

Sqlite不支持ALTER TABLE ADD CONSTRAINT-因此，当您使用Sqlite时，应省略对create_foreign_key（）的附加调用。

文档中有一个明确的注释：

由于SQLite对更改表的支持有限，因此在创建表后不能将外键约束添加到表中。

Peewee循环外键依赖项异常

1 个答案: