Steve Rodriguez wrote:
> Hey guys n gals,
>
> New to python, having some problems with while loops, I would like to make
> a program quick once q or Q is typed, but thus far I can only get the
> first variable to be recognized. My code looks like:
>
> message = raw_input("-> ")
> while message != 'q':
> s.send(message)
> data = s.recv(2048)
> print str(data)
> message = raw_input("-> ")
> s.close()
> print("Shutting Down")
>
> I've tried:
>
> while message != 'q' or 'Q':
This evaluates
(message != "q") or "Q"
and "Q" is always True in a boolean context:
>>> bool("Q")
True
> while message != 'q' or message != 'Q':
This is true when at least one of the subexpressions
message != "q"
message != "Q"
is True. When the value of message is "q" it must be unequal to "Q" and vice
versa, so there is always at least one true subexpression.
> while message != ('q' or 'Q'):
When you try the right side in the interactive interpreter you get
>>> "q" or "Q"
'q'
so this is the same as just
message != "q"
> Any ideas would be much appreciated! Thanks! :D
while message.lower() != "q":
while message not in ("q", "Q"):
while message != "q" and message != "Q":
There's another one that chains boolean expressions
while "q" != message != "Q":
that I don't recommend here but that is very readable for checking number
intervals:
if 0 <= some_number < 1:
print some_number, "is in the half-open interval [0,1)"
You also may consider an infinite loop:
while True:
message = raw_input("-> ")
if message in ("q", "Q"):
break
s.send(message)
...
This avoids the duplicate raw_input().
PS: You sometimes see
message in "qQ"
but this is buggy as it is true when the message is either
"q", "Q", or "qQ".
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