如何在Flask-restless中返回“已存在”错误？

Question

我想为异常做一些处理程序。我在python中使用了Flask-Flask和SQLAlchemy的组合。

我的问题是：

当我使用DB中已经存在的对象向api发送请求时，SQLAlchemy显示异常：

IntegrityError: (IntegrityError) column <column_name> is not unique u'INSERT INTO ...

因此，我尝试在create_api方法中添加属性validation_exceptions：

manager.create_api( ... , validation_exceptions=[IntegrityError])

但是响应json包含：

{
    "validation_errors": "Could not determine specific validation errors"
} 

服务端api显示异常：

Traceback (most recent call last):
  File "C:\Python27\lib\site-packages\flask_restless\views.py", line 797, in _extract_error_messages
    left, right = str(exception).rsplit(':', 1)
ValueError: need more than 1 value to unpack

Flask-restless中的异常验证不适用于此类异常(IntegrityError)

我该怎么办？是否可以为异常创建一些处理程序，并在json中返回我自己的错误消息？

Answer 1

documentation (v0.17.0到本文发布之日)声明：

目前，Flask-Restless期望指定的验证错误的实例将具有错误属性，该属性是将字段名映射到错误描述的字典(注意:每个字段一个错误)。

因此，要更改validation_errors的内容，异常需要一个包含字典的errors属性。此字典的内容将在服务器响应中显示为validation_errors。

来自flask-restless/tests/test_validation.py

class TestSimpleValidation(ManagerTestBase):
"""Tests for validation errors raised by the SQLAlchemy's simple built-in
validation.
For more information about this functionality, see the documentation for
:func:`sqlalchemy.orm.validates`.
"""

def setup(self):
    """Create APIs for the validated models."""
    super(TestSimpleValidation, self).setup()

    class Person(self.Base):
        __tablename__ = 'person'
        id = Column(Integer, primary_key=True)
        age = Column(Integer, nullable=False)

        @validates('age')
        def validate_age(self, key, number):
            if not 0 <= number <= 150:
                exception = CoolValidationError()
                exception.errors = dict(age='Must be between 0 and 150')
                raise exception
            return number

        @validates('articles')
        def validate_articles(self, key, article):
            if article.title is not None and len(article.title) == 0:
                exception = CoolValidationError()
                exception.errors = {'articles': 'empty title not allowed'}
                raise exception
            return article

    class Article(self.Base):
        __tablename__ = 'article'
        id = Column(Integer, primary_key=True)
        title = Column(Unicode)
        author_id = Column(Integer, ForeignKey('person.id'))
        author = relationship('Person', backref=backref('articles'))

    self.Article = Article
    self.Person = Person
    self.Base.metadata.create_all()
    self.manager.create_api(Article)
    self.manager.create_api(Person, methods=['POST', 'PATCH'],
                            validation_exceptions=[CoolValidationError])

请求：

data = dict(data=dict(type='person', age=-1))
response = self.app.post('/api/person', data=dumps(data))

响应：

HTTP/1.1 400 Bad Request

{ "validation_errors":
    {
      "age": "Must be between 0 and 150",
    }
}

Answer 2

您可以使用预处理器来捕获验证错误。

def validation_preprocessor(data, *args, **kwargs):
    # validate data by any of your cool-validation-frameworks
    if errors:
        raise ProcessingException(description='Something went wrong', code=400)

manager.create_api(
    Model,
    methods=['POST'],
    preprocessors=dict(
        POST=[validation_preprocessor]
    )
)

但我不确定这是不是一个好方法。