关于Flask和导入模块的困惑

Question

我有一个烧瓶应用程序,我正在尝试组织并遵循适当的烧瓶文件夹结构和模块/导入已在许多教程中显示.

我真的不知道为什么我要做某些事情,这从来都不是一件好事.

我的烧瓶应用程序布局如下:

/steam
    run.py
    /steamapp
        __init__.py
        config.py
        tasks.py
        views.py
        /static
           css.css
        /templates
           template.html

run.py

from steamapp import app
app.run()

init .py

from celery import Celery
from config import secrets, constants
from flask.ext.openid import OpenID
from flask import Flask
import praw

def make_celery(app):
    celery = Celery(app.import_name, backend='amqp', broker='amqp://guest@localhost//')
    celery.conf.update(app.config)
    TaskBase = celery.Task
    class ContextTask(TaskBase):
        abstract = True
        def __call__(self, *args, **kwargs):
            with app.app_context():
                return TaskBase.__call__(self, *args, **kwargs)
    celery.Task = ContextTask
    return celery


app = Flask(__name__)
oid = OpenID(app)
app.debug = True
app.secret_key = secrets.APP_SECRET_KEY
celery = make_celery(app)


reddit = praw.Reddit(constants.USERAGENT)
reddit.login(constants.USERNAME, secrets.PASSWORD)

rs = praw.Reddit(constants.USERAGENT)
rs.set_oauth_app_info(secrets.CLIENT_ID, secrets.CLIENT_SECRET, constants.REDIRECT_URL)


from steamapp import views
from steamapp import tasks

views.py

from steamapp import app, oid, celery
from tasks import get_auth_url, check_reddit_oauth, request_steam_api
from flask import Flask, session, redirect, request, render_template, url_for
from config import secrets, constants
from flask.ext.openid import OpenID
import os
import praw
import time
import json
import string
import random
import requests


def id_generator(size=6, chars=string.ascii_uppercase + string.digits):
    return ''.join(random.choice(chars) for _ in range(size))


def check_session_id(session):
    if "id" in session:
        return True
    else:
        return render_template("error.html", error=["?", "No session ID found."])


@app.route("/register")
def register():
    if "id" not in session:
        session["id"] = id_generator()
    authorize_url = get_auth_url(session)
    return redirect(authorize_url)


@app.route("/redirect/")
@oid.loginhandler
def redirect_to_steam_oauth():
   ...
   ...
   ...
   etc

tasks.py

from steamapp import celery, rs
from config import secrets, constants
import requests

@celery.task
def get_auth_url(session):
    return rs.get_authorize_url(session["id"], scope="identity")

当我运行run.py时,我得到以下回溯:

Traceback (most recent call last):
  File "/home/andy/Desktop/Python Projects/Finished/steam/run.py", line 1, in <module>
    from steamapp import app
  File "/home/andy/Desktop/Python Projects/Finished/steam/steamapp/__init__.py", line 25, in <module>
    from steamapp import views
  File "/home/andy/Desktop/Python Projects/Finished/steam/steamapp/views.py", line 20, in <module>
    @app.route("/register")
NameError: name 'app' is not defined

我无法理解整体上发生了什么 - 我需要解释为什么我正在做我正在做的事情:

1. 分离app.run()到从steamapp导入应用程序的单独python文件的原因是什么？为什么app.run()不在views.py中？

2. 什么是init .py实际上意味着要做什么？为什么在views.py和tasks.py导入呢？

3. 为什么它在导入时没有定义"app"的追溯__init__.py？

感谢能够提供一些澄清的任何人,因为我已经失去了对正在发生的事情的感觉.

---

编辑:

我已经更新了OP中的代码,并且在阅读了@ dreamriver的回复之后我做了更改.

我仍然有点困惑 - 为什么我在views.py中导入flask时,它已经在init .py中导入,其中导入了视图并且可能是从中运行的？

此外,芹菜工人如何与此相关？将tasks.py设置为Celery工作程序将返回以下回溯:

 celery -A tasks worker --loglevel=info
Traceback (most recent call last):
  File "/usr/local/bin/celery", line 11, in <module>
    sys.exit(main())
  File "/usr/local/lib/python2.7/dist-packages/celery/__main__.py", line 30, in main
    main()
  File "/usr/local/lib/python2.7/dist-packages/celery/bin/celery.py", line 81, in main
    cmd.execute_from_commandline(argv)
  File "/usr/local/lib/python2.7/dist-packages/celery/bin/celery.py", line 769, in execute_from_commandline
    super(CeleryCommand, self).execute_from_commandline(argv)))
  File "/usr/local/lib/python2.7/dist-packages/celery/bin/base.py", line 305, in execute_from_commandline
    argv = self.setup_app_from_commandline(argv)
  File "/usr/local/lib/python2.7/dist-packages/celery/bin/base.py", line 465, in setup_app_from_commandline
    self.app = self.find_app(app)
  File "/usr/local/lib/python2.7/dist-packages/celery/bin/base.py", line 485, in find_app
    return find_app(app, symbol_by_name=self.symbol_by_name)
  File "/usr/local/lib/python2.7/dist-packages/celery/app/utils.py", line 229, in find_app
    sym = symbol_by_name(app, imp=imp)
  File "/usr/local/lib/python2.7/dist-packages/celery/bin/base.py", line 488, in symbol_by_name
    return symbol_by_name(name, imp=imp)
  File "/usr/local/lib/python2.7/dist-packages/kombu/utils/__init__.py", line 92, in symbol_by_name
    module = imp(module_name, package=package, **kwargs)
  File "/usr/local/lib/python2.7/dist-packages/celery/utils/imports.py", line 101, in import_from_cwd
    return imp(module, package=package)
  File "/usr/lib/python2.7/importlib/__init__.py", line 37, in import_module
    __import__(name)
  File "/home/andy/Desktop/Python Projects/Finished/steam/steamapp/tasks.py", line 1, in <module>
    from steamapp import celery, rs
ImportError: No module named steamapp

Answer 1

您必须将app明确导入到您的视图文件中.错误说应用程序未在views.py中定义,这是因为您需要执行以下操作:

from steamapp import app

位于视图文件的顶部.

在__init__.py将您的目录到Python包.在解释器启动时,__init__.py添加了$ PYTHONPATH上的所有目录,sys.path这使得它们可以导入.__init__.py导入时定义的模块也会执行.这使得它非常有用,例如导出代码的主api.如果你查看了烧瓶的源代码,你会看到你所做的所有代码

from flask import Flask, request etc

实际上是在单独的文件中以较小的功能块定义,然后将有趣的部分公开__init__.py.

如前所述, Python不希望包中的模块成为启动文件.Python包装有点混乱.这个 stackoverflow的答案帮助我理解了其中的一些问题,即相对导入完全中断,因为它们是在直接执行文件时__name__设置的,'__main__'而在通过导入使用时设置文件本身.

除了几个模块级别的全局变量之外__name__,__package__没有任何内容显式地导入到顶级命名空间.这就是为什么app隐藏在views.py文件中无法使用的原因.

这会回答你的问题吗？

编辑

您需要导入views.py __init__.py,否则您的views.py文件将不会被执行,并且您的路由都不会被定义.您需要将app导入views.py,因为app不在您的views.py文件能够访问的命名空间中.这两个文件相互导入的模式称为循环导入,可能很棘手,但这里很好.您应该知道,在加载模块后,它会被缓存,因此在再次导入模块时不会重新执行.

您的芹菜问题在我看来,sys.path当芹菜启动时,python没有看到您的应用程序.显示sys.path那时的输出会很有帮助.我的猜测是,如果将工作目录添加到$ PYTHONPATH,问题将得到解决.当您使用pip等安装东西时,包将被添加到python默认知道如何找到它们的地方.