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Date: Tue, 08 Nov 2011 18:15:40 +0100
From: Jean-Michel Pichavant <jeanmichel@sequans.com>
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To: Gnarlodious <gnarlodious@gmail.com>
Subject: Re: Help catching error message
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Xref: x330-a1.tempe.blueboxinc.net comp.lang.python:15471

Gnarlodious wrote:
> What I say is this:
>
> def SaveEvents(self,events):
>    try:
>       plistlib.writePlist(events, self.path+'/Data/Events.plist') #
> None if OK
>    except IOError:
>       return "IOError: [Errno 13] Apache can't write Events.plist
> file"
>
> Note that success returns"None" while failure returns a string.
>
> I catch the error like this:
>
> errorStatus=Data.Dict.SaveEvents(Data.Plist.Events)
> if errorStatus: content=errorStatus
>
> It works, but isn there a more elegant way to do it? As in, one line?
> I can imagine if success returned nothing then content would remain
> unchanged. Isn't there a built-in way to send an error string back and
> then catch it as a variable name?
>
> This is Py3 inside WSGI.
>
> -- Gnarlie
>   
Hi,

There's no need to rephrase an exception unless you want to *add* 
information.

def saveEvents(self,events):
    plistlib.writePlist(events, self.path+'/Data/Events.plist') 

try:   
    Data.Dict.SaveEvents(Data.Plist.Events)
except IOError, exc: # i'm using python 2.5, this except clause may have 
changed in py3
    content = str(exc)

JM