I'm trying open a txt file and read it, but I'm getting a type error and I'm unsure why. If you you could please provide a reasoning along with the correct syntax, I'm trying to get a better grasp of what's going on underneath. Here's the code, it's pretty simple I think:
from sys import argv
script = argv
filename = argv
txt = open(filename)
print "Here's your file %r" %filename
print txt.read()
Muchas Gracias
argv is a list, not a string. You want
script = argv[0]
filename = argv[1]
Consider using argparse instead of handing sys.argv directly:
>>> import argparse
>>> parser = argparse.ArgumentParser(description="Print a file.")
>>> parser.add_argument("path", type=str, nargs=1, help="Path to the file to be printed.")
_StoreAction(option_strings=[], dest='path', nargs=1, const=None, default=None, type=<type 'str'>, choices=None, help='Path to the file to be printed.', metavar=None)
>>> args = parser.parse_args()
>>> print args
Namespace(path=[<...>])
It looks much more complicated, but it will make your command-line utility much more flexible, easier to expand, and it will ensure you get proper documentation on the command line.
First of all, argv is a list of arguments. Open doesn't take a list. That's why you're getting a type error.
Second, open (should) take 2 parameters, the filename and the mode (yes mode is optional, but get in the habit of putting it there. Replace with
import sys
script = sys.argv[0]
filename = sys.argv[1]
txt = open(filename, 'r')
print "Here's your file %r" %filename
print txt.read()
argv will be a list, while filename should be a string.
You should probably be using filename = argv[1]
