[Tutor] How do you make a module

[alan.gauld@bt.com]

> I'm stumped about how to handle a namespace issue.
> 
> In the __main__ program I open a file for output like this:
> 
>     DEBUGOUT = open('f:/source/linkextr2/debug_rr.txt', 'w')
> 
> and the functions that __main__ calls output logging data to 
> it.  Now that those functions have been moved to their own 
> module (crawler.py), the functions have no idea of DEBUGOUT's 
> existence.  

from crawler import DEBUGOUT

should make it available. Alternatively set a local variable 
to it via the module:

dbgout = crawler.DEBUGOUT

> to pass DEBUGOUT to crawl_site(), but crawl_site calls other functions
> within the same module and those functions also need to 
> access the DEBUGOUT file.

That means they are using the name as a global variable which is 
bad practice from a purist point of view. Better to have all 
functions that use it take it as an argument. However that 
may be a big edit job so use the techniques above to fake the 
global...

> Is there a way to make DEBUGOUT created in __main__ available 
> to the other module (crawler.py)?

Ah, I see, its the other way arround.

OK, In that case modify crawl_site() to take the DEBUGOUT as 
an argument and at the begginning set a module level variable 
in crawler.py to the log file

Like so:

if __name__ == "__main__":
    DBGOUT = open(....)
    crawl_site(DBGOUT,....)

And in crawler.py

DEBUGOUT = None  # initialise the module level variable

def crawl_site(dbgfile,....):
   global DEBUGOUT
   DEBUGOUT = dbgfile # now set variable with value from __main__ 
   .....as before....

def crawl_another():
   # blah blah
   DEBUGOUT.write("OOPS, An error happened\n")

etc etc...


HTH,

Alan g.