Question

“python manage.py runserver”在下面的情况下运行。它不运行main.py或app包。

烧瓶

myApp          
    main.py

bash> python main.py

flask_script

manage.py
myApp
    __init__.py
    main.py

bash> python manage.py runserver

manage.py

from flask_script import Manager
from myApp import app

manager = Manager(app)

if __name__ == "__main__":
    manager.run()

main.py

from flask import Flask
app = Flask(__name__)  

@app.route('/')
def hello():   
    return "Hello World"

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

且__init__.py为空

Answer 1

我回答自己的问题，因为每当我在这个热情友好的社区中提出任何要求时，我都会得到任意的弊端。

我发现“一种方式”可能不是“方式”。

文件结构

|manage.py
|myApp
  |__init__.py
  |main.py
  |blog
    |__init__.py
    |blog.py

<强> manage.py

from flask_script import Manager
from flask import Flask

from myApp.main import interface as main_inaterface
from myApp.blog.blog import interface as blog_interface

app.register_bluprint(main_interface, url_prefix='/main')
app.register_bluprint(blog_interface, url_prefix='/blog')

manager = Manager(app)

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

myApp .__ init __。py

from flask import Flask
app = Flask(__name__)

import myApp

main.py

from flask import Blueprint

interface = Blueprint('myApp', __name__)

@interface.route('/')
def hello():
    return "from main"

<强>博客

from flask import Blueprint

interface = Blueprint('blog', __name__)

@interface.route('/')
def blog():
  return "From Blog"

烧瓶脚本文件结构

1 个答案: