如何为 python flask 应用程序提供结构

Question

我是 python 烧瓶的新手

在单个文件中使用 MongoDB 试验一些端点，如下所示

from flask import Flask, request
from flask.ext.mongoalchemy import MongoAlchemy
app = Flask(__name__)
app.config['DEBUG'] = True
app.config['MONGOALCHEMY_DATABASE'] = 'library'
db = MongoAlchemy(app)


class Author(db.Document):
    name = db.StringField()


class Book(db.Document):
    title = db.StringField()
    author = db.DocumentField(Author)
    year = db.IntField();


@app.route('/author/new')
def new_author():
    """Creates a new author by a giving name (via GET parameter)
    e.g.: GET /author/new?name=Francisco creates a author named Francisco
    """
    author = Author(name=request.args.get('name', ''))
    author.save()
    return 'Saved :)'






@app.route('/authors/')
def list_authors():
    """List all authors.

    e.g.: GET /authors"""
    authors = Author.query.all()
    content = '<p>Authors:</p>'
    for author in authors:
        content += '<p>%s</p>' % author.name
    return content

if __name__ == '__main__':
    app.run()

以上代码包含指向 post 的两个端点并获取工作正常的数据

知道正在寻找一种方法将代码分离到不同的文件中，例如

数据库连接相关的代码应该在不同的文件中

from flask import Flask, request
from flask.ext.mongoalchemy import MongoAlchemy
app = Flask(__name__)
app.config['DEBUG'] = True
app.config['MONGOALCHEMY_DATABASE'] = 'library'
db = MongoAlchemy(app)

我应该能够在定义和使用架构的不同文件中获取数据库引用

class Author(db.Document):
    name = db.StringField()


class Book(db.Document):
    title = db.StringField()
    author = db.DocumentField(Author)
    year = db.IntField();

和路线将是不同的文件

@app.route('/author/new')
def new_author():
    """Creates a new author by a giving name (via GET parameter)
    e.g.: GET /author/new?name=Francisco creates a author named Francisco
    """
    author = Author(name=request.args.get('name', ''))
    author.save()
    return 'Saved :)'






@app.route('/authors/')
def list_authors():
    """List all authors.

    e.g.: GET /authors"""
    authors = Author.query.all()
    content = '<p>Authors:</p>'
    for author in authors:
        content += '<p>%s</p>' % author.name
    return content

我应该在端点文件中获取数据库模式的引用，请帮助我获取此结构

给我一些可以帮助我做的可以理解的示例或视频，我是 python 以及烧瓶的新手，请指出一些示例并帮助我了解更多，谢谢

Answer 1

基本结构可能如下所示：

/yourapp  
    /run.py  
    /config.py  
    /yourapp  
        /__init__.py
        /views.py  
        /models.py  
        /static/  
            /main.css
        /templates/  
            /base.html  
    /requirements.txt  
    /venv

应用于您的示例，它看起来像这样。

run.py: 开始申请。

from yourapp import create_app

app = create_app()
if __name__ == '__main__':
    app.run()

config.py：包含配置，可以添加子类来区分开发配置、测试配置和生产配置

class Config:
    DEBUG = True
    MONGOALCHEMY_DATABASE = 'library'

yourapp/_init_.py：初始化您的应用程序，创建一个 Flask 实例。（也使您的应用程序成为一个包）。

from flask import Flask
from flask.ext.mongoalchemy import MongoAlchemy
from config import Config

db = MongoAlchemy()

def create_app():
    app = Flask(__name__)
    app.config.from_object(Config)

    db.init_app(app)
    
    from views import author_bp
    app.register_blueprint(author_bp)

    return app

yourapp/models.py: 包含你的不同型号。

from . import db

class Author(db.Document):
    name = db.StringField()


class Book(db.Document):
    title = db.StringField()
    author = db.DocumentField(Author)
    year = db.IntField();

yourapp/views.py：有时也称为 routes.py。包含您的 url 端点和相关行为。

from flask import Blueprint
from .models import Author

author_bp = Blueprint('author', __name__)

@author_bp.route('/author/new')
def new_author():
    """Creates a new author by a giving name (via GET parameter)
    e.g.: GET /author/new?name=Francisco creates a author named Francisco
    """
    author = Author(name=request.args.get('name', ''))
    author.save()
    return 'Saved :)'

@author_bp.route('/authors/')
def list_authors():
   """List all authors.     
   e.g.: GET /authors"""
   authors = Author.query.all()
   content = '<p>Authors:</p>'
   for author in authors:
       content += '<p>%s</p>' % author.name
   return content

yourapp/static/... 包含您的静态文件。

yourapp/templates/.. 包含您的模板。

requirements.txt 有您的包依赖项的快照。

venv (Virtualenv) 文件夹，您的 python 库将能够在包含的环境中工作。

参考文献：

Have a look at this related question.

Good example of a widely used project structure.

如何为 python flask 应用程序提供结构

how to give structure to python flask application

python

mongodb

flask

mongoalchemy