An integer is required? open()

Question

I have a very simple python script that should scan a text file, which contains lines formatted as id='value' and put them into a dict. the python module is called chval.py and the input file is in.txt. here's the code:

import os,sys
from os import *
from sys import *

vals = {}

f = open(sys.argv[1], 'r')

for line in val_f:
    t = line.split('=')
    t[1].strip('\'')
    vals.append(t[0], t[1])

print vals

f.close()

when i try to run it i get:

Traceback (most recent call last):
File "chval.py", line 9, in ? f = open(sys.argv[1], 'r') TypeError: an integer is required

I'm using python 2.4... because i've been challenged to not use anything newer, is there something about open() that I don't know about? Why does it want an integer?

anything after that line is untested. in short: why is it giving me the error and how do i fix it?

Answer 1

Because you did from os import *, you are (accidenally) using os.open, which indeed requires an integer flag instead of a textual "r" or "w". Take out that line and you'll get past that error.

Answer 2

Don't do import * from wherever without a good reason (and there aren't many).

Your code is picking up the os.open() function instead of the built-in open() function. If you really want to use os.open(), do import os then call os.open(....). Whichever open you want to call, read the documentation about what arguments it requires.

Answer 3

Also of note is that starting with Python 2.6 the built-in function open() is now an alias for the io.open() function. It was even considered removing the built-in open() in Python 3 and requiring the usage of io.open, in order to avoid accidental namespace collisions resulting from things such as "from blah import *". In Python 2.6+ you can write (and can also consider this style to be good practice):

import io
filehandle = io.open(sys.argv[1], 'r')

Answer 4

Providing these parameters resolved my issue:

with open('tomorrow.txt', mode='w', encoding='UTF-8', errors='strict', buffering=1) as file:
    file.write(result)

Answer 5

From http://www.tutorialspoint.com/python/os_open.htm you could also keep your import and use

file = os.open( "foo.txt", mode )

and the mode could be :

os.O_RDONLY: open for reading only
os.O_WRONLY: open for writing only
os.O_RDWR : open for reading and writing
os.O_NONBLOCK: do not block on open
os.O_APPEND: append on each write
os.O_CREAT: create file if it does not exist
os.O_TRUNC: truncate size to 0
os.O_EXCL: error if create and file exists
os.O_SHLOCK: atomically obtain a shared lock
os.O_EXLOCK: atomically obtain an exclusive lock
os.O_DIRECT: eliminate or reduce cache effects
os.O_FSYNC : synchronous writes
os.O_NOFOLLOW: do not follow symlinks