没有此类文件或目录的相对路径 [英] No such file or directory for relative path

问题描述

我在一个python文件-backend.py中有一个Bottle应用程序。该文件包含以下定义：

I have a bottle application in one python file - backend.py. The file contains these definitions:

variable = {
    'field': [f for f in csv.DictReader(open('../data/fields.csv', 'rb'), delimiter=';')]
}

def run_fcgi():
    from bottle import FlupFCGIServer
    run(port=8080, server=FlupFCGIServer)

if __name__ == "__main__":
    run(host='0.0.0.0', port=8087, server='waitress')

当我运行此应用程序时，例如：

When I run this app like:

python backend.py

应用程序是成功启动。

当我由主管作为fcgi应用程序（fcgi.py）运行此应用程序时：

When I run this application as fcgi application (fcgi.py) by supervisor:

#!my_path_to_python

if __name__ == '__main__':
    import backend
    backend.run_fcgi()

我有一个错误：

Traceback (most recent call last):
  File "path_to_my_project/fcgi.py", line 9, in <module>
    import backend
  File "path_to_my_project/backend.py", line 49, in <module>
    'msk': [i for i in csv.DictReader(open('../data/fields.csv', 'rb'), delimiter=';')],
IOError: [Errno 2] No such file or directory: '../data/fields.csv'

任何想法？

推荐答案

我认为最好不要依赖工作目录。并使用相对于使用此路径的文件的路径。我的意思是您可以随时计算路径：

I think It's better not to rely on working directory. and use path relative to file that is uses this path. I mean you can calculate path on the fly:

import os
csv_path = '../data/fields.csv'
csv_path = os.path.join(os.path.dirname(__file__), csv_path)

在这种情况下，您可以在不同的环境上运行脚本。完整路径将用于确保您不依赖于工作目录。

In this case one be able to run your script on different environment. an full path will be used to be sure that You don't depends on working directory.

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