Passing an argument to a python script and opening a file

Question

I've built a script that takes a filename as an argument and extracts all lines that match a certain pattern. Trouble is I can't open the filename - I keep getting :

"TypeError: coercing to unicode: need string or buffer"

It's complaining about the line info = open(name, 'r').

Here's the code:

import re
import sys
print sys.argv[1:]

keyword = 'queued='
pattern = re.compile(keyword)

name = sys.argv[1:]
inf = open(name, 'r') 
outf = open("test.txt", 'w')

for line in inf:
    if pattern.search(line):
        outf.write(line)

And I call it with

`extract.py trunc.log`

Any ideas what I'm doing wrong?

Answer 1

sys.argv[1:] is a list, not a string. When you slice a list, you get a list back -- even if you only take 1 element with the slice. You need to give open a string. Perhaps you wanted sys.argv[-1] (the last element)?

As a side note, the python standard library provides commandline parsing options -- One is the excellent argparse module which was introduced in python 2.7, but can be installed with very minimal effort on older python versions (I use it with python2.6 regularly).

Answer 2

This:

name = sys.argv[1:]

makes name into a list of all arguments except the first. You mean:

name = sys.argv[1]

For real-world usage, you might want to look into argparse.