迭代 sqlalchemy 模型定义的列的方法? [英] method of iterating over sqlalchemy model's defined columns?

问题描述

我一直在尝试弄清楚如何迭代 SQLAlchemy 模型中定义的列列表.我想要它为几个模型编写一些序列化和复制方法.我不能只是迭代 obj.__dict__ 因为它包含很多 SA 特定项目.

I've been trying to figure out how to iterate over the list of columns defined in a SQLAlchemy model. I want it for writing some serialization and copy methods to a couple of models. I can't just iterate over the obj.__dict__ since it contains a lot of SA specific items.

有谁知道从下面获取id 和desc 名称的方法?

Anyone know of a way to just get the id and desc names from the following?

class JobStatus(Base):
    __tablename__ = 'jobstatus'

    id = Column(Integer, primary_key=True)
    desc = Column(Unicode(20))

在这个小案例中，我可以轻松创建一个:

In this small case I could easily create a:

def logme(self):
    return {'id': self.id, 'desc': self.desc}

但我更喜欢自动生成 dict 的东西(对于较大的对象).

but I'd prefer something that auto-generates the dict (for larger objects).

推荐答案

您可以使用以下功能:

def __unicode__(self):
    return "[%s(%s)]" % (self.__class__.__name__, ', '.join('%s=%s' % (k, self.__dict__[k]) for k in sorted(self.__dict__) if '_sa_' != k[:4]))

它会排除 SA magic 属性，但不会排除关系.所以基本上它可能会加载依赖项，父母，孩子等，这绝对是不可取的.

It will exclude SA magic attributes, but will not exclude the relations. So basically it might load the dependencies, parents, children etc, which is definitely not desirable.

但实际上要容易得多，因为如果你从 Base 继承，你有一个 __table__ 属性，这样你就可以:

But it is actually much easier because if you inherit from Base, you have a __table__ attribute, so that you can do:

for c in JobStatus.__table__.columns:
    print c

for c in JobStatus.__table__.foreign_keys:
    print c

参见如何从SQLAlchemy 映射对象 - 类似问题.

Mike 请参阅Mapper.c 和 Mapper.mapped_table.如果使用 0.8 及更高版本，请参阅 Mapper.attrs 和相关函数.

Edit by Mike: Please see functions such as Mapper.c and Mapper.mapped_table. If using 0.8 and higher also see Mapper.attrs and related functions.

Mapper.attrs<的示例/a>:

Example for Mapper.attrs:

from sqlalchemy import inspect
mapper = inspect(JobStatus)
for column in mapper.attrs:
    print column.key

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