Mat*_*ock 4 python python-module peewee
假设我有几个简单的模型位于food.py
:
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()
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后来,我有了food.py
和food.db
。但是假设Taco
和Spaghetti
模型变得庞大且复杂,所以我想将它们分成不同的文件。具体来说,我想food
在我的文件夹中创建一个PYTHONPATH
具有典型层次结构的文件夹:
food/
- __init__.py
- BaseModel.py
- Taco.py
- Spaghetti.py
- db/
- food.db
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我想将模型放入各自的.py
文件中,并有一个__init__.py
如下所示的文件:
import peewee as pw
db = pw.SqliteDatabase('./db/food.db')
from . import BaseModel
from . import Taco
from . import Spaghetti
db.connect()
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但是,这显然不起作用,因为BaseModel.py
无法访问db
. 如果可以通过这种方式模块化多个 peewee 模型,那么正确的方法是什么?
显然,诀窍是连接到BaseModel.py
文件中的数据库。我将给出模块内容的完整概述。假设顶级文件夹已命名food
并位于PYTHONPATH
. 最后假设food.db
存在food/db/food.db
并且已被填充(例如,如问题中第一个代码块的底部)。
以下是模块文件:
__init__.py
from Taco import Taco
from Spaghetti import Spaghetti
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基础模型.py
import peewee as pw
db = pw.SqliteDatabase('/abs/path/to/food/db/food.db')
class BaseModel(pw.Model):
class Meta:
database = db
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玉米饼
import peewee as pw
from BaseModel import BaseModel
class Taco(BaseModel):
has_cheese = pw.BooleanField()
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意大利面.py
import peewee as pw
from BaseModel import BaseModel
class Spaghetti(BaseModel):
has_meatballs = pw.BooleanField()
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例如,现在您可以编写一个脚本(当然位于模块文件夹之外),例如:
主要.py
import food
for t in food.Taco.select():
print "Taco", t.id, ("has" if t.has_cheese else "doesn't have"), "cheese"
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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
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