How do I convert the str to int without an error

Question

def grading():
if student_score <40:
print("Grade: F")
elif student_score <46:
print("Grade: E")
elif student_score <50:
print("Grade: D")
elif student_score <60:
print("Grade: C")
elif student_score <70:
print("Grade: B")
elif student_score >70:
print("Grade: A")

student_score=int(input("Enter the student score or enter to exist"))
total_score=0
number_of_students=0
grading()

while student_score != "":
total_score+= student_score
number_of_students+=1
student_score=int(input("Enter the student_score or enter to exist"))

grading ()
print("number of students =",number_of_students)
average= total_score / number_of_students
print("Average=",average)

Answer 1

Doesn't this do what you want?

https://code.sololearn.com/c58VoL8eadrW/?ref=app

Answer 2

You're closing the while loop when the input is "" (nothing), but you try and cast the input to int before checking that there is anything in the string.

score = input()
while score != "":
score = int(score)
# do_stuff
score = input()

Answer 3

I think you might have done this.
Still, you can see the following code:-
s=int(input())
print(s)

Answer 4

You get invalid literal error, when the input isn't an integer.

In your while loop, you are comparing student_score with a string and it isn't a string...

Answer 5

AKSHAY I know. I am currently using pydroid3

Answer 6

Guys help I still can't correct the error anyone to help

Answer 7

Russ thanks very much it runs just fine..I will compare it to mine and make the necessary adjustments. Thanks

Answer 8

Yeah I did that it keeps telling me
ValueError: invalid literal for int() with base 10: ''

Answer 9

Adeniran not sure but I think this may be due sololearn input problem as your program needs more than 1 input and it requires all input to be given in starting at time of running, if you are not giving proper input then it will cause error..

Answer 10

Hi

Answer 11

Use type casting property
Example:-
Store="58"
restore=int(store)
print(restore)