different behaviour for user defined exception with attribute args

Hi all For an exception defined as below class OptionError(Exception): def __init__(self, args): self.args = args def __str__(self): return repr(self.v) an iteration is happening when the exception is raised Meanwhile for almost the same structured exception replacing th

"TypeError: 'int' object is not callable"

when I was executing the below code I got "TypeError: 'int' object is not callable" exception. Why is it so? if type(c) == type(ERROR): c can be a string or an integer representing an error -- Thanks & Regards visco -- http://mail.python.org/mailman/listinfo/python-list

Any adv. in importing a module and some objects in the same module, into the same file?

http://docs.python.org/library/logging.html While going thr' the above link i came across import statements "import logging import logging.handlers" What is the use of second import as the first import will be enough(AFAIK) to access anything intended by the second import? Is there any kind of a

how to know the importing file name from an imported file?

Hi Is there a way to know the name of the script(say A), which is importing a module(say B), from B? ie in above situation i should be able to get name 'A' through some way in B, when A contains an 'import B' statement. -- Thanks & Regards visco -- http://mail.python.org/mailman/listinfo/python-

Re: to get name of file opened

First of all Thanks Dave for the reply On Sat, 2009-03-28 at 09:51 -0500, Dave Angel wrote: > First question is why you need os.open(), and not the open() function. > I'll guess that you need some of the access modes (e.g. for file > sharing) that you get from the low level functions. So assumi

difference between os.fdopen and builtin open

Hi I was wondering the difference between os.fdopen()(or os.open() not considering the difference in args) and builtin open(). Can anyone help me? -- Thanks & Regards visco -- http://mail.python.org/mailman/listinfo/python-list

to get name of file opened

Hi Is there any way to get the name of the file opened from the file object 'f' which i get through the code f = os.fdopen(os.open("trial', os.O_WRONLY|os.O_CREAT), "w") The situation will be like i can access only the above variable 'f'. f.name is having '' instead of filename 'trial' Or if n

subprocess module: execution of standard binaries without shell?

hi all while getting used to with subprocess module i failed in executuing a) but succeeded in running b). Can anyone explain me why as i am providing absolute path? Is this has to do anything with shared library.. which must be accessed based on system variables? a) pipe = subprocess.Popen("/bi