Flask:如何在应用程序根目录中读取文件? [英] Flask: How to read a file in application root?

问题描述

我的Flask应用程序结构如下

My Flask application structure looks like

application_top/
         application/
                    static/
                          english_words.txt
                    templates/
                             main.html
                     urls.py
                     views.py
         runserver.py

当我运行runserver.py时，它将在localhost:5000处启动服务器. 在我的views.py中，我尝试以

When I run the runserver.py, it starts the server at localhost:5000. In my views.py, I try to open the file english.txt as

f = open('/static/english.txt')

它给出错误IOError: No such file or directory

如何访问此文件?

推荐答案

我认为问题是您将/放在了路径中.删除/，因为static与views.py处于同一级别.

I think the issue is you put / in the path. Remove / because static is at the same level as views.py.

我建议将settings.py的级别设置为与views.py相同的级别，或者许多Flask用户更喜欢使用__init__.py，但我不这样做.

I suggest making a settings.py the same level as views.py Or many Flask users prefer to use __init__.py but I don't.

application_top/
    application/
          static/
              english_words.txt
          templates/
              main.html
          urls.py
          views.py
          settings.py
    runserver.py

如果这是您要设置的方式，请尝试以下操作:

If this is how you would set up, try this:

#settings.py
import os
# __file__ refers to the file settings.py 
APP_ROOT = os.path.dirname(os.path.abspath(__file__))   # refers to application_top
APP_STATIC = os.path.join(APP_ROOT, 'static')

现在，在您的视图中，您只需执行以下操作即可:

Now in your views, you can simply do:

import os
from settings import APP_STATIC
with open(os.path.join(APP_STATIC, 'english_words.txt')) as f:
    f.read()

根据您的要求调整路径和水平.

Adjust the path and level based on your requirement.

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