How do I check the users input in Python3?

Question

So my code says this:

a = input("Choose a letter")

if a == "A" or "B" or "D":
cr = "false" 

elif a == "C":
  cr = "true"

if cr == "false":
 print("Sorry you did not asnwer correctly")
 import sys
 sys.exit()

elif cr == "true":
 print("Well done!")

But for some reason, whatever I put, it says it is wrong. Any help? I am new to python and have come from Lua.

"it is wrong", what do you mean to say. is program throwing error or showing incorrect output — angrysumit, Oct 02 '18 at 06:32
Please try to provide the code as you wrote it - it is very badly indented and does not run this way. — Christian König, Oct 02 '18 at 06:37
To be a little bit more pythonic you should use this: `if a in ("A", "B", "C") : ...`. — Sven-Eric Krüger, Oct 02 '18 at 06:37

score 3 · Accepted Answer · edited Oct 02 '18 at 06:45

3

if a == "A" or "B" or "D": this evaluates to TRUE always. Because "B" or "D" always evaluates to TRUE everytime. So this might boil down to if 0 or 1 or 1 no matter what.

To fix this you need compare after every conditional operator. if a == "A" or a == "B" or a == "D":

edited Oct 02 '18 at 06:45

timgeb

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answered Oct 02 '18 at 06:31

Vineeth Sai

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Red Cricket · Answer 2 · 2018-10-02T06:40:36.920

2

if a == "A" or "B" or "D":
   cr = "false"

Is wrong. You want:

if a == "A" or a == "B" or a == "D":
   cr = "false"

One nice thing you could do is this:

if a in [ 'A', 'B', 'D' ]:
   cr = "false"

Or even better rewrite your script to this:

a = input("Choose a letter")

if a == 'C':
 print("Well done!")
else:
 print("Sorry you did not asnwer correctly")
 import sys
 sys.exit()

edited Oct 02 '18 at 06:40

answered Oct 02 '18 at 06:32

Red Cricket

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`if a in {'A', 'B', 'D'}: ` is better. – Austin Oct 02 '18 at 06:37
1

@Austin not necessary, since `BUILD_SET` takes time, too. I'd use tuples. – bereal Oct 02 '18 at 06:42

How do I check the users input in Python3?

2 Answers2