I need to realize a complex if-elif-else statement in Python but I don't get it working.
The elif line I need has to check a variable for this conditions:
80, 443 or 1024-65535 inclusive
I tried
if
...
# several checks
...
elif (var1 > 65535) or ((var1 < 1024) and (var1 != 80) and (var1 != 443)):
# fail
else
...
This should do it:
elif var == 80 or var == 443 or 1024 <= var <= 65535:
It's often easier to think in the positive sense, and wrap it in a not:
elif not (var1 == 80 or var1 == 443 or (1024 <= var1 <= 65535)):
# fail
You could of course also go all out and be a bit more object-oriented:
class PortValidator(object):
@staticmethod
def port_allowed(p):
if p == 80: return True
if p == 443: return True
if 1024 <= p <= 65535: return True
return False
# ...
elif not PortValidator.port_allowed(var1):
# fail
1 < a < 5 gets converted to (1 < a) < 5 which is always true, since (1 < a) is a bool with value of 0 or 1, and is always less than 5. It's nice that Python provides more human-readable range syntax!I think the most pythonic way to do this for me, will be
elif var in [80,443] + range(1024,65535):
although it could take a little time and memory (it's generating numbers from 1024 to 65535). If there's a problem with that, I'll do:
elif 1024 <= var <= 65535 or var in [80,443]:
if
...
# several checks
...
elif not (1024<=var<=65535 or var == 80 or var == 443)
# fail
else
...
It is possible to write like this:
elif var1 in [80, 443] or 1024 < var1 < 65535
This way you check if var1 appears in that a list, then you make just 1 check, didn't repeat the "var1" one extra time, and looks clear:
if var1 in [80, 443] or 1024 < var1 < 65535:
print 'good'
else:
print 'bad'
....:
good
if
...
# several checks
...
elif ((var1 > 65535) or ((var1 < 1024)) and (var1 != 80) and (var1 != 443)):
# fail
else
...
You missed a parenthesis.
