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Peewee循环外键依赖项异常

[英]Peewee circular foreign key dependencies exception

I'm trying to replicate Peewee example app using circular dependences as explained in the docs , despite the creator clearly states this is usually a bad idea. 我试图按照文档中的解释使用循环依赖来复制Peewee示例应用程序,尽管创建者明确指出这通常是个坏主意。 This is the code mainly copied from the docs: 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()

I'm getting the an exception in the line commented with #Error. 我在用#Error注释的行中得到了一个异常。 This line is needed, as explained in the docs: 如文档中所述,这行是必需的:

When you call create_table we will again encounter the same issue. 当您调用create_table时,我们将再次遇到相同的问题。 For this reason peewee will not automatically create a foreign key constraint for any deferred foreign keys. 因此,peewee不会为任何延迟的外键自动创建外键约束。

To create the tables and the foreign-key constraint, you can use the SchemaManager.create_foreign_key() method to create the constraint after creating the tables. 要创建表和外键约束,可以在创建表后使用SchemaManager.create_foreign_key()方法创建约束。

This is the exception I get using Python 3.5.2: 这是我使用Python 3.5.2获得的例外:

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

Sqlite does not support ALTER TABLE ADD CONSTRAINT -- so when you're using Sqlite you should omit the additional call to create_foreign_key(). 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。

There is a clear NOTE in the docs that says: 在文档中有一个明确的注释:

Because SQLite has limited support for altering tables, foreign-key constraints cannot be added to a table after it has been created. 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。

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