Weird Behaviour In Python If Clause
Solution 1:
user == "paper" or "Paper"
is always true. The or
operator tests the expressions on either side of itself, and if either is true, the result of the or
is also true. Your test above checks (up to) two things:
- Is the expression
user == "paper"
true? If so, the whole expression is true, so don't check the second part, becauseTrue or x
is always true regardless of the value ofx
. - Is the expression
"Paper"
true? And because non-zero-length strings are true in Python, this part is always true.
So even if the first part is false, the second part is always true, so the expression as a whole is always true.
You wanted something like this:
user == "paper" or user == "Paper"
or, better yet:
user in("paper", "Paper")
or, best of all:
user.lower() == "paper"
Solution 2:
You can also do this with lists and in
:
if user in ["paper", "Paper"]:
paper()
or using regex:
import re
user='paper'
if re.match('papers?', user):
paper()
elif re.match('[Rr]ock', user):
rock()
with regexes you can also do case-insensitive match:
importreuser='paper'if re.match('papers?', user, re.I):
paper()
which will match all: paper, PapER, PaperS, ...
Solution 3:
you want:
ifuser== "paper"oruser== "Paper":
Same for the others as well.
If you just put
if"Paper":
Python evaluates it as if this_value_is_true
. Same with the code you have basically evaluates to "if user variable equals 'paper' or True" which would always be tue.
Solution 4:
I believe I know where your problem comes from:
ifuser== "paper"oruser== "Paper":
That should fix the problem
Solution 5:
This:
ifuser== "paper" or "Paper":
… is parsed as this:
if (user== "paper") or "Paper":
If user
actually is equal to "paper"
, that's if True or "Paper"
, which is True
.
Otherwise, that's if False or "Paper"
, which is "Paper"
.
Since True
and "Paper"
are both truthy, the if
always happens.
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