Question

Python: Create new function dedicated to file open?

Not sure if it's worth it but...

from sys import argv

new_file = argv

def creation(new_one):
new_one = open(new_file, 'w')
(Would I need to return new_one?)

creation(target)

While attempting run, I'll get the error, name 'target' is not defined.

What would I need to do to fix this to successfully create a function for file opening?

I have no idea how to define target...

I could do file i/o but that's only from not using functions.

Any hints would be greatly appreciated.

Answer 1

I was thinking of defining target as the name of the txt file I plan on creating. Will try that after I get back from running.

Any input still welcome.

Answer 2

A simple example in python to open a file
file_name=str(raw_input("Enter file name:"))
if not file_name.endwiths(".txt"):
file_name=file_name + ".txt"
def creation(new_file):
fp=open(new_file,"w")
fp.write("i have opened the file")
fp.close()

creation(file_name)