How can I open a text file and assign it's contents to a variable in a Python function?

Question

I have a text file called inputs.txt thats in a folder called task1 the input file contains a series of strings that I need to process in a Python function.

I need to write a function that can open this intputs.txt file and assign the the string contents to the a variable S

So far I've got:

def open_func(task1/test-input.txt):   # syntax error thrown here with forward-slash
    S = open(task1/test-input.txt, "r")
    print(S)
    return S

But this throws a syntax error at the forward-slash

The input file currently contains acbcbba, which I want to be passed to the variable S

What am I doing wrong?

EDIT:

I've attached a screen shot of the solution I've tried but I'm still getting the "no file or directory test-input.txt" error

Cheers

Answer 1

There are multiple problems in here:

1. What is inside the parenthesis in the definition must be a parameter, not a string (so replace that task1/test-input.txt by something like file or filename , because task1/test-input.txt is what you are trying to open, not a parameter to the function). OR

2. If you want to open a file named task1/test-input.txt you need to surround it by quotes (either simple or double, I personnaly prefer double), so "task1/test-input.txt"

3. the open function opens a file handle , not the content of the file. You need to call read() on the handle, then close() it. So something like:

    file = open(filename, "r") S = file.read() file.close() print(S) return S

4. Also, you should use the with syntax as pointed out in the comment, which simplifies the above to (since that automatically close s the handle):

    with open(filename, "r") as file: S = file.read() print(S) return S

Answer 2

You need to use a filename variable to pass to the function. To do that you declare a variable with its value encapsulated by quotes like this:

def open_func(filename):
    f = open(filename, "r")
    content = f.read()
    f.close()
    print(content)
    return content

path = "task1/test-input.txt"
content = open_func(path)
# do something with the file content now

Regarding EDIT: The file you are opening needs to be in an accessible path from where you run your script. So if your folder structure looks like this:

task1/
    script.py
    test-input.txt

You need to call from this path, if you call your script from within "task1/":

path = "test-input.txt

To get your working directory, you can use this snippet to find out:

import os
print(os.getcwd())