On 20 February 2013 17:54, <mikp...@gmail.com> wrote:
> On Tuesday, February 19, 2013 5:47:16 PM UTC, Michael Torrie wrote:
>> On 02/19/2013 02:24 AM, mikp...@gmail.com wrote:
>>
>> > Or rather: what would you try to catch in this particular case?
>>
>>
>> As Peter said, nothing for now. But you seem very resistant to telling
>>
>> us what exception was raised.
>
>
> Michael believe me:
> I am not resistant or try to hide anything!
> As written before, I don't know what exception to search for, so I wrote the
> (wrong) code:
> except:
> print "error"
> Let's why I don't have a clue about it.
> But someone already explained me that I should not do this.
You don't need to look for errors. If you remove the try/except then
they show up automatically. For example (in the interpreter):
$ python
Python 2.7.3 (default, Sep 26 2012, 21:51:14)
[GCC 4.7.2] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> open('Desktop')
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
IOError: [Errno 21] Is a directory: 'Desktop'
The last three lines above are the error message that people were
expecting you to show here. They contains lots of useful information:
1) The type of the error
2) A message "Is a directory" and in this case a cryptic code 21 (that
some might find useful).
3) The line of code that caused the error (this is more useful when
running code saved in a file).
What you are doing, however, is this:
>>> try:
... open('Desktop')
... except:
... print('An error occurred...')
...
An error occurred...
Which gives a much less useful error message. So just remove the try/except.
Oscar
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