[SOLVED] Python Phone Number Validator

https://code.sololearn.com/cA3A20a15A2a You can find answer in the above link.

import re pattern = r"^[189][0-9]{7}$" if re.match(pattern, input()): print("Valid") else: print("Invalid") =========================================== I hope this helps you all. ^ is for starts with [189], [0-9] digits for {7} repetiotions $ to end total of [189] + [0-9]{7} which is equals to 8. Hope you have understood my answer😊

import re a = input() pattern = r"[189]" if re.match(pattern,a): if len(a) == 8: print("Valid") else: print("Invalid") else: print("Invalid")

import re #your code goes here number = input() pattern = r"^(1|8|9)" match = re.match(pattern, number) if match and len(number) == 8: print("Valid") else: print("Invalid")

import re #your code goes here number = input() if re.match(r"^[1|8|9]\d{7}$", number): print("Valid") else: print("Invalid")

import re pattern = r"^[189][0-9]{7}" str = input() match = re.search(pattern, str) if match: print("Valid") else: print("Invalid") What wrong with this code?

i still get one case not correct this is my code where is the mistake ? import re #your code goes here pattern = r"^1|8|9 + [0-9]{7}$" num = str(input()) if re.search(pattern, num): print("Valid") else: print("Invalid")

Most probably you're getting this failure in last test case cause re.search() searches for the whole string even if the string contains multi-lines and tries to find a match of the substring in all the lines of string but you want your numeric number to only START with 1,8 or 9. better to use re.match() instead

Here is my answer to that guys: import re number = input() pattern = r"^[189][0-9]" match = re.match(pattern, number) if len(number) == 8 and match: print("Valid") else: print("Invalid")

@Arya Huda Arrasyid You need a $ at the end of your regex, otherwise, the phone number can start with the correct 8 digits but then go on indefinitely. It can only be 8 digits

I recommend to test the code on the SL Playground with a few data inputs and a few print statements and see what happend and solve that. Sp you could see that 01234567 gives a valid number, what unfortunately is not true from task description. The solution for it will be to code that the number has to begin with 1 or 8 or 9.

import re num = input () pattern = r"^[1|8|9][0-9]*$" if re.match(pattern,num) and len(num)==8: print("Valid") else: print("Invalid")

I think this is simplest: import re phone = input() pattern = r"^[189][0-9]{7}$" if re.match(pattern, phone): print("Valid") else: print("Invalid")

import re #tu código va aquí pattern = "\d\1|8|9{1,9}$" number = input() if re.search(pattern, number) and len(number) == 8: print("Valid") else: print("Invalid")

Try this: import re #your code goes here str = input() pattern = r"^(1|8|9)+\d{7,7}$" if re.match(pattern , str): print("Valid") else: print("Invalid")

import re print("Valid" if re.match(r'[789]\d{7}$', input()) else 'Invalid')

Okime Obia It's the python library for regular expressions

Mi respuesta es: import re pattern = r"(1|8|9)[0-9]{7}$" entrada = input() if re.match(pattern, entrada): print("Valid") else: print("Invalid")

import re num = input () pattern = r"^[1|8|9][0-9]*$" if re.match(pattern,num) and len(num)==8: print("Valid") else: print("Invalid")

import re x=input() if re.fullmatch("[189]\d{7}",x): print('Valid') else: print("Invalid")