I am trying to open a file and read from it and if the file is not there, I catch the exception and throw an error to stderr. The code that I have:
for x in l:
try:
f = open(x,'r')
except IOError:
print >> sys.stderr, "No such file" , x
but nothing is being printed to stderr, does open create a new file if the file name doesn't exist or is the problem somewhere else?
Try this:
from __future__ import print_statement
import sys
if os.path.exists(x):
with open(x, 'r') as f:
# Do Stuff with file
else:
print("No such file '{}'".format(x), file=sys.stderr)
The goal here is to be as clear as possible about what is happening. We first check if the file exists by calling os.path.exists(x). This returns True or False, allowing us to simply use it in an if statement.
From there you can open the file for reading, or handle exiting as you like. Using the Python3 style print function allows you to explicitly declare where your output goes, in this case to stderr.
open() call inside of a try/except block is almost certainly better here -- the call to open can fail for reasons besides the file not existing (what happens in your code if the file exists, but the current process doesn't have read permission for it? BOOM -- an exception that will need to be caught.) This is a case where EAFP beats LBYL.You have the os.path.exists function:
import os.path
os.path.exists(file_path)
returns bool
It works for me. Why can't you make use of os.path.exists()
for x in l:
if not os.path.exists(x):
print >> sys.stderr , "No such file", x
open(x, 'r')without enclosing it in a try statement, do you get an error?open('asdfghj'), do you get an error? Assuming you don't have such a file with that name, of course.