What is causing the error (index out of range)-CodePudding

def main():
    plate = input("Plate: ")
    if is_valid(plate):
        print("Valid")
    else:
        print("Invalid")


def is_valid(s):
    index = []
    for i in s:
        if i.isdigit():
            index  = i
            break
    print(index)
    if 6 >= len(s) >= 2 and s[0:1].isalpha() and s.isupper() and index[0] != '0':
        return True
main()

Before I added and index[0] != '0' the code worked perfectly, but for some reason after adding that piece of code, when I go to input "KEVIN" an error(index out of range) pops up. How do I prevent this error from popping while still checking out the requirements for the code in the if statement?

CodePudding user response：

A smaller example shows the problem

def is_valid(s):
    index = []
    for i in s:
        if i.isdigit():
            index  = i
            break
    print(index)
    if 6 >= len(s) >= 2 and s[0:1].isalpha() and s.isupper() and index[0] != '0':
        return True

is_valid("KEVIN")

"KEVIN" doesn't contain any digits, i.isdigit() is never True and the index list remains empty. Add a check for that case.

def is_valid(s):
    index = []
    for i in s:
        if i.isdigit():
            index  = i
            break
    print(index)
    if (6 >= len(s) >= 2 and s[0:1].isalpha() and s.isupper() 
            and index and index[0] != '0'):
        return True

CodePudding user response：

The code does not consider the edge case for when the string s has no digits. In that case, index will be an empty list and index[0] != '0' will throw an error. Consider adding a condition to check for the length of s.