Re: Begniner Question

[bruno modulix]

Glen wrote:
#!/usr/local/bin/python
import sys
print "1.\tDo Something"
print "2.\tDo Something"
print "3.\tDo Something"
print "4.\tDo Something"
print "5.\tDo Something"
print "6.\tExit"
choice=raw_input("Please Enter Choice: ")
if int(choice)==1:
   print "Here"
else:
   pass
if int(choice)==2:
else:
   pass
(snip)
File "./Code.py", line 20
   else:
      ^
IndentationError: expeted an indented block
What am I doing wrong?
Err... I'm not totally sure, but I think it could be possible that you 
perhaps should insert an indented block between lines 19 and 20 !-)

Thank-you
You're welcome !-)
A bit more seriously, now... the construct:
if some_condition:
else:
  pass
is syntactically incorrect. Python expects an indented block between the 
'if' line and the 'else' line. If you don't have the needed code yet, 
you need to put a 'pass' here :

if some_condition:
  pass
else:
  pass
This will of course do nothing else than useless tests, but it will at 
least satisfy the language's rules.

--
bruno desthuilliers
python -c "print '@'.join(['.'.join([w[::-1] for w in p.split('.')]) for 
p in '[EMAIL PROTECTED]'.split('@')])"
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