Catching an exception while using a Python 'with' statement


Question

To my shame, I can't figure out how to handle exception for python 'with' statement. If I have a code:

with open("a.txt") as f:
    print f.readlines()

I really want to handle 'file not found exception' in order to do somehing. But I can't write

with open("a.txt") as f:
    print f.readlines()
except:
    print 'oops'

and can't write

with open("a.txt") as f:
    print f.readlines()
else:
    print 'oops'

enclosing 'with' in a try/except statement doesn't work else: exception is not raised. What can I do in order to process failure inside 'with' statement in a Pythonic way?

1
250
4/5/2009 9:32:06 PM

Accepted Answer

from __future__ import with_statement

try:
    with open( "a.txt" ) as f :
        print f.readlines()
except EnvironmentError: # parent of IOError, OSError *and* WindowsError where available
    print 'oops'

If you want different handling for errors from the open call vs the working code you could do:

try:
    f = open('foo.txt')
except IOError:
    print('error')
else:
    with f:
        print f.readlines()
225
3/8/2011 4:36:12 PM

The best "Pythonic" way to do this, exploiting the with statement, is listed as Example #6 in PEP 343, which gives the background of the statement.

@contextmanager
def opened_w_error(filename, mode="r"):
    try:
        f = open(filename, mode)
    except IOError, err:
        yield None, err
    else:
        try:
            yield f, None
        finally:
            f.close()

Used as follows:

with opened_w_error("/etc/passwd", "a") as (f, err):
    if err:
        print "IOError:", err
    else:
        f.write("guido::0:0::/:/bin/sh\n")

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