如何实现工厂类？-Python问题

How to implement a factory class?(如何实现工厂类？)

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问题描述

我希望能够基于枚举类创建对象，并使用字典。大概是这样的：

class IngredientType(Enum):
    SPAM        = auto() # Some spam
    BAKE_BEANS  = auto() # Baked beans
    EGG         = auto() # Fried egg

class Ingredient(object):
    pass    
class Spam(Ingredient):
    pass
class BakedBeans(Ingredient):
    pass
class Egg(Ingredient):
    pass


class IngredientFactory(object):
    """Factory makes ingredients"""

    choice = {
        IngredientType.SPAM: IngredientFactory.MakeSpam,
        IngredientType.BAKED_BEANS: IngredientFactory.MakeBakedBeans,
        IngredientType.EGG: IngredientFactory.MakeEgg
    }

    @staticmethod
    def make(type):
        method = choice[type]
        return method()

    @staticmethod
    def makeSpam():
        return Spam()

    @staticmethod
    def makeBakedBeans():
        return BakedBeans()

    @staticmethod
    def makeEgg():
        return Egg()

但我收到错误：

NameError: name 'IngredientFactory' is not defined

由于某种原因，无法创建词典。我在哪里做错了？

推荐答案

查看Bruce Eckel's book后，我想到了这个：

#Based on Bruce Eckel's book Python 3 example
# A simple static factory method.
from __future__ import generators
import random
from enum import Enum, auto

class ShapeType(Enum):
    CIRCLE  = auto() # Some circles
    SQUARE  = auto() # some squares

class Shape(object):
    pass

class Circle(Shape):
    def draw(self): print("Circle.draw")
    def erase(self): print("Circle.erase")

class Square(Shape):
    def draw(self): print("Square.draw")
    def erase(self): print("Square.erase")

class ShapeFactory(object):

    @staticmethod
    def create(type):
        #return eval(type + "()") # simple alternative
        if type in ShapeFactory.choice:
            return ShapeFactory.choice[type]()

        assert 0, "Bad shape creation: " + type    

    choice = { ShapeType.CIRCLE:  Circle,
               ShapeType.SQUARE:  Square                
             }

# Test factory
# Generate shape name strings:
def shapeNameGen(n):

    types = list(ShapeType)

    for i in range(n):
        yield random.choice(types)

shapes = 
  [ ShapeFactory.create(i) for i in shapeNameGen(7)]

for shape in shapes:
    shape.draw()
    shape.erase()

这会让用户从枚举中选择类类型，并阻止任何其他类型。这也意味着用户不太可能写出拼写错误的"坏字符串"。他们只使用枚举。然后，测试的输出如下所示：

Circle.draw
Circle.erase
Circle.draw
Circle.erase
Square.draw
Square.erase
Square.draw
Square.erase
Circle.draw
Circle.erase
Circle.draw
Circle.erase
Square.draw
Square.erase

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