Simple python if/else syntax not working

Question

I am trying to create a simple test-scorer that grades your test and gives you a response - but a simple if/else function isn't running -

Python -

testScore = input("Please enter your test score")
if testScore <= 50:
 print "You didn't pass... sorry!" 
elif testScore >=60 and <=71:
 print "You passed, but you can do better!"

The Error is -

Traceback (most recent call last):
 File "python", line 6
 elif testScore >= 60 and <= 71:
 ^
SyntaxError: invalid syntax

• Jim Wright: What is the error?
• Ma0: Python is high-level but not straight-up English.. elif testScore >=60 and <=71: should be modified into this elif 60 <= testScore <= 71:
• user8739422: It works now. Thanks!

Answer 1

You missed testScore in elif statement

 testScore = input("Please enter your test score")
if testScore <= 50:
 print "You didn't pass... sorry!" 
elif testScore >=60 and testScore<=71:
 print "You passed, but you can do better!"

Answer 2

The below shown way would be the better way of solving it, you always need to make the type conversion to integer when you are comparing/checking with numbers.

input() in python would generally take as string

 testScore = input("Please enter your test score")
 if int(testScore) <= 50:
 print("You didn't pass... sorry!" )
 elif int(testScore) >=60 and int(testScore)<=71:
 print("You passed, but you can do better!")

Answer 3

You made some mistakes here:

• You are comparing a string with an Integer if testScore <= 50:

• You have missed the variable here --> elif testScore >=60 and <=71:

I think those should be like this --->

• if int(testScore) <= 50:
• elif testScore >=60 and testScore<=71:

And try this, it is working --->

testScore = input("Please enter your test score")
if int(testScore) <= 50:
 print ("You didn't pass... sorry!") 
elif testScore >=60 and testScore<=71:
 print ("You passed, but you can do better!")