为什么Python open（）函数不接受“目录/文件名”作为参数？

Question

Local environment: Python 3, Bottle, MacOs

Remote environment: Python 3, Bottle, Pythonanywhere

This works in my local environment but not in my remote environment:

@route('/test')
def test():
    '''Function tests file open issue.'''
    with open('uploads/Project2.csv', 'r', newline='') as file:
        content = ""
        content = file.read()
    return content

This works in my remote environment but not in my local environment:

@route('/test')
def test():
    '''Function tests file open issue.'''
    with open('uploads', 'r', newline='') as file:
        content = ""
        content = file.read()
    return content

In the first case, I pass a file path to the open function. If I pass a folder name to it returns this error:

IsADirectoryError: [Errno 21] Is a directory: 'uploads'

In the second case, I pass a folder name to the open function. If I pass a file path it returns error:

NotADirectoryError: [Errno 20] Not a directory: 'uploads/Project2.csv'

I am baffled. Any ideas?

Answer 1

First you have to be certain whether the path exists or not on your remote server.

import os 
os.path.exists(<your path>)

second, you dont have to declare your content variable, you can just declare it like this.

content = file.read()

Third,

"uploads" is a directory not a file. Provide a file name in your
directory like you have provided in your local environment. if 
"upload" is not a subdirectory of your code directory, then provide
absolute path. like 
upload = "/home/ubuntu/env/uploads/projects.csv"

Answer 2

In both environments "uploads" is a subdirectory of the code directory but...

In the local environment a relative path was sufficient:

"uploads/file"

In the remote environment an absolute path was required:

"/home/my_projects/project/uploads/file"

I think that this is to do with Bottle being a WSGI object in the remote environment but not in the local environment.