Python: Error message parsing

Question

I'm trying to handle IOError generated when trying to open a non existent file. I do:

try:
    inputFile = open('nosuchfile', 'r')
except IOError as exception:
    print 'error: %s' % (exception)

This gives me:

error: [Errno 2] No such file or directory: 'nosuchfile'

But I'm only interested in the message and not the [Errno 2] part. So I change my code.

print 'error: %s' % (exception.strerror)

But now I get:

error: No such file or directory

Where did the name of the file go? I know I could just print the file name separately, but I would really like how (if at all) the name was stored in the exception, but not in either of its arguments (printing exception.errno gives 2).

I am using version 2.7.3.

phihag · Accepted Answer · 2012-09-30 19:14:45Z

5

The filename is stored in, well, the filename property:

try:
    inputFile = open('nosuchfile', 'r')
except IOError as exception:
    print ('error: %s: %r' % (exception.strerror, exception.filename))

answered Sep 30, 2012 at 19:14

phihag

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6 Comments

Jayanth Koushik Over a year ago

Oh got it. But why is this not revealed when I print exception.args?

Mr_and_Mrs_D Over a year ago

NB: exception.strerror is not an Exception attribute

phihag Over a year ago

@Mr_and_Mrs_D Well, it does work, so why do you think it wouldn't be an attribute?

Mr_and_Mrs_D Over a year ago

It is an IOError attribute - not an Exception attribute - apparently one must call str(exception)

phihag Over a year ago

@Mr_and_Mrs_D str(exception) is the whole message including the filename, which is not being asked for. strerror is just the core error string. And of course strerror and filename are only available in IOError exception objects. But that's not a problem here, since we only catch IOError in the first place.

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