如何在Elixir/SqlAlchemy中创建类属性和类属性

Question

我有一个类：

from sys import stderr
from elixir import *
from types import *

class User(Entity):
    using_options(tablename="users")
    first_name = Field(String(50))
    middle_name = Field(String(50))
    last_name = Field(String(50))

    def __get_name__ (self):
        first_name = self.first_name if self.first_name is not None else ""
        middle_name = self.middle_name if self.middle_name is not None else ""
        last_name = self.last_name if self.last_name is not None else ""
        return " ".join((first_name, middle_name, last_name)).strip()

    def __set_name__ (self,string):
        first_name = ""
        middle_name = ""
        last_name = ""
        split_string = string.split(' ')
        if len(split_string) == 1:
            first_name = string
        elif len(split_string) == 2:
            first_name, last_name = split_string
        elif len(split_string) == 3:
            first_name, middle_name, last_name = split_string
        else: #len(split_string) > 3:
            first_name = split_string[0]
            last_name = split_string[-1]
            middle_name = " ".join(split_string[1:-2])
        self.first_name = first_name
        self.middle_name = middle_name
        self.last_name = last_name

    name = property(__get_name__,__set_name__)

我想运行一个查询，像这样：

def get_user(user):
    found = None
    if type(user) in [IntType,StringType]:
        if type(user) is StringType:
            where = or_(User.first_name==user,
                        User.middle_name==user,
                        User.last_name==user,
                        User.name==user)
            qry = User.query.filter(where)
        elif type(user) is IntType:
            where = or_(User.id==user,
                        User.employee_id==user)
            qry = User.query.filter(where)
        try:
            found = qry.one()
        except NoResultFound:
            print >> stderr, "Couldn't find '%s'" % user
    elif type(user) == User:
        found=user
    return found

但是，生成的SQL查询看起来像这样：

SELECT users.first_name AS users_first_name, 
       users.middle_name AS users_middle_name, 
       users.last_name AS users_last_name
FROM users 
WHERE users.first_name = 'Joseph'
   OR users.middle_name = 'M'
   OR users.last_name = 'Schmoe'
   OR false

注意到这里的false代替了User.name字段。

我遇到了这个错误：

sqlalchemy.exc.OperationalError: (OperationalError) no such column: false 

我想要的SQL查询应该是这样的：

SELECT users.name
FROM users 
WHERE users.name = 'Joseph M Schmoe'

补充：我之前的第二个SQL查询不太正确，因为我真正想要的是一种被动的方式，在数据库中创建一个'name'字段，这个字段是由'first_name'、'middle_name'和'last_name'拼接而成的。

补充2：我觉得下面这个方法几乎可以满足我的需求。不过，我还是在寻找合适的表达方式。

补充3：看起来这个方法能满足我的需求。所以我把它作为答案放在这里。

Answer 1

你定义了一个属性，这个属性在对象被赋值后就可以使用了。

注意：我为了方便修改了你的代码

from sqlalchemy.ext.declarative import declarative_base
from sqlalchemy import Column, Integer, DateTime, String
from datetime import datetime

Base = declarative_base()
class User(Base):
    first_name = Column('first_name', String)
    middle_name = Column('middle_name', String)
    last_name = Column('last_name', String)

    def __get_name__ (self):
        first_name = self.first_name if self.first_name is not None else ""
        middle_name = self.middle_name if self.middle_name is not None else ""
        last_name = self.last_name if self.last_name is not None else ""
        return " ".join((first_name, middle_name, last_name)).strip()

    def __set_name__ (self,string):
        first_name = ""
        middle_name = ""
        last_name = ""
        split_string = string.split(' ')
        if len(split_string) == 1:
            first_name = string
        elif len(split_string) == 2:
            first_name, last_name = split_string
        elif len(split_string) == 3:
            first_name, middle_name, last_name = split_string
        else: #len(split_string) > 3:
            first_name = split_string[0]
            last_name = split_string[-1]
            middle_name = " ".join(split_string[1:-2])
        self.first_name = first_name
        self.middle_name = middle_name
        self.last_name = last_name

    name = property(__get_name__,__set_name__)

所以 first_name, middle_name 和 last_name 是类的属性。当你定义一个属性时，需要有这个对象的实例。

In [13]: User.first_name
Out[13]: Column('first_name', String(), )

In [14]: User.name
Out[14]: <property object at 0x26fef70>

在上面的例子中，你可以看到区别。在你设置这个属性或者属性的任何字段之前，它总是会是空的。

你必须给它赋值

In [16]: u1 = User()

In [17]: u1.name = "first middle last"

In [18]: u1.name
Out[18]: 'first middle last'

In [19]: u1.first_name
Out[19]: 'first'

之后你就可以使用这个属性了。

这将帮助你理解你的问题。你的属性是和实例关联的，不能通过类来访问它。

Answer 2

from sqlalchemy.ext.hybrid import hybrid_property

@hybrid_property
def name (self):
    first_name = self.first_name if self.first_name is not None else ""
    middle_name = self.middle_name if self.middle_name is not None else ""
    last_name = self.last_name if self.last_name is not None else ""
    return " ".join((first_name, middle_name, last_name)).strip()

@name.setter
def name (self,string):
    first_name = ""
    middle_name = ""
    last_name = ""
    split_string = string.split(' ')
    if len(split_string) == 1:
        first_name = string
    elif len(split_string) == 2:
        first_name, last_name = split_string
    elif len(split_string) == 3:
        first_name, middle_name, last_name = split_string
    else: #len(split_string) > 3:
        first_name = split_string[0]
        last_name = split_string[-1]
        middle_name = " ".join(split_string[1:-2])
    self.first_name = first_name
    self.middle_name = middle_name
    self.last_name = last_name

这里是表达式的部分：

@name.expression
def name (cls):
    f = cls.first_name
    m = cls.middle_name
    l = cls.last_name
    return f+' '+m+' '+l