Question

假设我有几个位于food.py的简单模型：

import peewee as pw

db = pw.SqliteDatabase('food.db')

class BaseModel(pw.Model):
    class Meta:
        database = db

class Taco(BaseModel):
    has_cheese = pw.BooleanField()

class Spaghetti(BaseModel):
    has_meatballs = pw.BooleanField()

db.connect()

# populate with some data if table doesn't exist
from random import random
if not Taco.table_exists():
    db.create_table(Taco)
    for _ in range(10):
        Taco.create( has_cheese = (random() < 0.5) )
    db.commit()
if not Spaghetti.table_exists():
    db.create_table(Spaghetti)
    for _ in range(10):
        Spaghetti.create( has_meatballs = (random() < 0.5) )
    db.commit()

之后，我有food.py和food.db。但是，让我们说Taco和Spaghetti模型变得庞大而复杂，所以我想将它们分成不同的文件。具体来说，我想在food中使用典型的层次结构创建一个PYTHONPATH文件夹：

food/
    - __init__.py
    - BaseModel.py
    - Taco.py
    - Spaghetti.py
    - db/
        - food.db

我想将模型放入各自的.py文件中，并拥有一个__init__.py文件，如下所示：

import peewee as pw

db = pw.SqliteDatabase('./db/food.db')

from . import BaseModel
from . import Taco
from . import Spaghetti

db.connect()

但是，这显然无效，因为BaseModel.py无法访问db。如果能够以这种方式模块化多个peewee模型，那么这样做的正确方法是什么？

Answer 1

显然，技巧是连接到BaseModel.py文件中的数据库。我将概述模块内容。假设顶级文件夹名为food并且位于PYTHONPATH中。最后假设food.db中存在food/db/food.db并且已填充（例如，如问题中第一个代码块的底部）。

以下是模块文件：

__初始化__。PY

from Taco import Taco
from Spaghetti import Spaghetti

BaseModel.py

import peewee as pw
db = pw.SqliteDatabase('/abs/path/to/food/db/food.db')

class BaseModel(pw.Model):
    class Meta:
        database = db

Taco.py

import peewee as pw
from BaseModel import BaseModel

class Taco(BaseModel):
    has_cheese = pw.BooleanField()

Spaghetti.py

import peewee as pw
from BaseModel import BaseModel

class Spaghetti(BaseModel):
    has_meatballs = pw.BooleanField()

现在，例如，您可以编写一个脚本（当然位于模块文件夹之外），例如：

main.py

import food

for t in food.Taco.select():
    print "Taco", t.id, ("has" if t.has_cheese else "doesn't have"), "cheese"

产生

Taco 1 has cheese
Taco 2 has cheese
Taco 3 has cheese
Taco 4 doesn't have cheese
Taco 5 doesn't have cheese
Taco 6 has cheese
Taco 7 has cheese
Taco 8 has cheese
Taco 9 doesn't have cheese
Taco 10 doesn't have cheese

Answer 2

您在路径中遇到问题：

__location__ = os.path.realpath(os.path.join(os.getcwd(), os.path.dirname(__file__)))
db = pw.SqliteDatabase(os.path.join(__location__, 'db/food.db'));

还尝试实现__init__个类并将db作为参数传递给它：

class BaseModel(pw.Model):
    def __init__(self, db = None)
        self.database = db

而不是__init__.py：

from BaseModel import BaseModel
db = pw.SqliteDatabase('./db/food.db')
bm = BaseModel(db)

Answer 3

有关使用peewee模块化烧瓶应用程序的说明，请参阅此文章：

http://charlesleifer.com/blog/structuring-flask-apps-a-how-to-for-those-coming-from-django/

模块化的peewee

3 个答案: