simple e-mail without regexp and without list

[Nykeriab]

Good evening. There's a question, I understand the point of error, but I can't find the right solution to the problem. The challenge is: Write an e-mail field holder using only the material that has been produced and the instruction that is in this task. Validator is a program that checks the correctness of the data. If the e-mail address is in place, remove YES, otherwise NO♪

Input data: email - string value

Output data: YES - if email is Validen, NO - unless email is Validen

Email field mask: @. - where ___ may contain a-z, A-Z, 0-9 and points. Each of the blocks shall contain at least one letter before the symbol @.

Regular expressions and lists cannot be usedI'm trying to set up a spontaneous rust and check that after @, the point is not the following symbol.

# Простой валидатор почтового адреса (без использования regual expression)
ввод строки пользователем
input_adr = input()
счетчик первого символа (проверяем есть ли в адресе хотя бы один символ)
inint_count = 0
верный адрес почты
correct_adr = 'alpaRist11@code.com'
поиск символов '.', '@' во входой строке
for i in input_adr:

at = '@'

dot = '.'
проверяем наличие первого символа в строке, которую ввел пользователь
до символа @, первый блок
for inint_count in input_adr:

if((ord(input_adr[inint_count])>= 48 and ord(input_adr[inint_count])<= 57) or

(ord(input_adr[inint_count])>=64 and ord(input_adr[inint_count])<=90)

or (ord(input_adr[inint_count])>= 97 and ord(input_adr[inint_count])<= 122)):

inint_count += 1
проверяем условия, что после символа @ идет хотя бы один символ,
что символы @ и . не идут вместе
if(inint_count > 0 and at > 0  and (dot-at) > 0 and (dot+1) < len(input_adr)):

print("Yes")

else:

print("No")

In error:

  File "main.py", line 20, in <module>

if((ord(input_adr[inint_count]>= 48) and ord(input_adr[inint_count]<= 57)) or

TypeError: string indices must be integers

[Jolied]

if((ord(input_adr[inint_count]>= 48) and ord(input_adr[inint_count]<= 57)) or 

Consider the brackets. You're getting the following.

ord(input_adr[inint_count]>= 48)

I mean, you. ord You're making it from a bell like that. ord(True) or ord(False)

Most likely you meant

ord(input_adr[inint_count]) >= 48

P. S.

And in fact, the code for the said rules, I would have done it like that:

address = 'test@A12.ru'

address_parts = address.split('@')
checked = address.count('@') == 1
проверить наличие точки
checked &= '.' in address_parts[1]
проверить наличие хотя бы одной буквы
checked &= any(letter.isalpha() for letter in address_parts[1])
проверить наличие букв, цифр и точки
checked &= all((letter.isalpha() or letter.isdigit() or letter == '.') for letter in address_parts[1])
ВНИМАНИЕ: в условии не сказано проверять блок до @, так что последнюю проверку можно удалить скорее всего
проверить наличие букв, цифр и точки
checked &= all((letter.isalpha() or letter.isdigit() or letter == '.') for letter in address_parts[0])
print(checked)

P.P.S.

If you can't use anything at all, you can use that code.

address = 'test@A14.ru'

checked = True
проверить кол-во символов @
symbols_count = 0
for i in address:

symbols_count += 1 if i == '@' else 0
checked &= symbols_count == 1
найти координату '@'
symbol_pos = 0
for pos in range(len(address)):

if address[pos] == '@':

symbol_pos = pos + 1

break
проверить наличие точки
isDot = False

for pos in range(symbol_pos, len(address)):

isDot |= address[pos] == '.'
checked &= isDot
проверить наличие хотя бы одной буквы
isAlpha = False

for pos in range(symbol_pos, len(address)):

isAlpha |= ord('a') <= ord(address[pos]) <= ord('z') or ord('A') <= ord(address[pos]) <= ord('Z')
checked &= isAlpha
проверить наличие букв, цифр и точки
isLetters = True

for pos in range(symbol_pos, len(address)):

isLetters &= ord('a') <= ord(address[pos]) <= ord('z') or ord('A') <= ord(address[pos]) <= ord('Z') or ord('0') <= ord(address[pos]) <= ord('9') or address[pos] == '.'
checked &= isLetters
print(checked)

P.P.P.S.

The coke which evaluates not only the part after @ but also the separate blocks between the points in this part:

address = '@a.com'

checked = True
проверить кол-во символов @
symbols_count = 0
for i in address:

symbols_count += 1 if i == '@' else 0
checked &= symbols_count == 1
найти координату '@'
symbol_pos = 0
for pos in range(len(address)):

if address[pos] == '@':

symbol_pos = pos + 1

break
проверить наличие точки
isDot = False

for pos in range(symbol_pos, len(address)):

isDot |= address[pos] == '.'
checked &= isDot
проанализировать блоки (расположенные между точками)
start = symbol_pos

for index in range(symbol_pos, len(address)):

if address[index] == '.' or index == len(address) - 1:

finish = index if address[index] == '.' else (index + 1)
    # проверить наличие хотя бы одной буквы
    isAlpha = False
    for pos in range(start, finish):
        isAlpha |= ord('a') <= ord(address[pos]) <= ord('z') or ord('A') <= ord(address[pos]) <= ord('Z')

    checked &= isAlpha

    # проверить наличие букв, цифр и точки
    isLetters = True
    for pos in range(start, finish):
        isLetters &= ord('a') <= ord(address[pos]) <= ord('z') or ord('A') <= ord(address[pos]) <= ord('Z') or ord('0') <= ord(address[pos]) <= ord('9')

    checked &= isLetters

    start = index + 1

print(checked)