Creating .exe file in Python
subhabrata.banerji at gmail.com
subhabrata.banerji at gmail.com
Tue Jun 16 12:23:08 EDT 2015
On Tuesday, June 16, 2015 at 9:33:58 PM UTC+5:30, Chris Angelico wrote:
> On Wed, Jun 17, 2015 at 1:17 AM, wrote:
> > Thanks. The scipy issue seems solved. But this silly issue is giving so much of time. I am checking. Please see a sample code,
> >
> > import sys
> > sys.stderr = sys.stdout
> > class Colors:
> > def Blue(self):
> > self.var="This is Blue"
> > print self.var
> > def Red(self):
> > print self.var
> >
> >
> >
> > if __name__ == "__main__":
> > Colors().Blue() #THIS IS FINE
> > Colors().Red() #NOT FINE
>
> You're still not saying what's going on. Did you try this code as a
> simple Python script first, before trying to bundle it up into an .exe
> file?
>
> Fortunately, my primary crystal ball is active, and I believe what's
> going on is that you expect Blue() to set something and then Red() to
> see it. However, you're calling those methods on two different
> throw-away objects, so they have separate state. What you expect to
> happen, I honestly have no idea. (Also, why are you fiddling with
> sys.stderr? You don't then appear to be using it, unless you have an
> issue with exceptions getting printed to the other stream.)
>
> ChrisA
Thank you for your reply. I am trying to experimenting over small ones before going to actual code. Your error interpretations are right.
I could work out a small solution please check if it is going fine.
It is giving me result.
if __name__ == "__main__":
c1=Colors()
c2=c1.Blue()
c3=c1.Red()
Regards,
Subhabrata Banerjee.
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