I want to print phone number validator program

Question

import re

def isValid(s):

# 1) Begins with 1,8 or 9

# 2) Then contains 7 or 8 or 9.

# 3) Then contains 8 digits

Pattern = re.compile("(1/8-9)?[8-9][0-9]{8}")

return Pattern.match(s)

# Driver Code

s = "347873923408"

if (isValid(s)):

print ("Valid")

else :

print ("Invalid")
#3/5 cases are correct

Answer 1

import re
#your code goes here
num = str(input())
pattern = "\A(1|8|9)\d{7}quot;

if re.match(pattern,num):
print("Valid")
else:
print("Invalid")

#I got my answer

Answer 2

import re
#your code goes here
num = str(input())
length = len(num)

pattern = r"^(1|8|9)"

if re.match(pattern,num) and length == 8:
print("Valid")
else:
print("Invalid")

# You can also do like this

Answer 3

Pattern = re.compile("[1|8|9][7-9][0-9]{8}")

'/' is a charecter there
'|' is works as or
So hope it works.

Answer 4

import re
#your code goes here
num = str(input())
pattern = "\A(1|8|9)\d{7}quot;

if re.match(pattern,num):
print("Valid")
else:
print("Invalid")

#answer

Answer 5

import re
#your code goes here
num = str(input())
pattern = "\A(1|8|9)\d{7}quot;

if re.match(pattern,num):
print("Valid")
else:
print("Invalid")

Answer 6

mport re

#your code goes here

pattern = "\A(1|8|9)\d{7}quot;
num_input = str(input())

match = re.match(pattern, num_input)

if match:
print("Valid")
else:
print("Invalid")

Answer 7

import re
#your code goes here
number = input()

pattern = r"[189]"

if re.match(pattern, number) and len(number)==8:
print("Valid")

else:
print("Invalid")

Answer 8

import re
#your code goes here
num = input()

pattern = r'^[1|8|9]'

match = re.findall(pattern,num)

if match and len(num)==8:
print("Valid")
else:
print("Invalid")

Answer 9

import re
#your code goes here
number = input()
pattern = r"(1|8|9)\d{7}quot;

if re.search(pattern,number):
print("Valid")
else:
print("Invalid")

Answer 10

Nooo it doesn't work

Answer 11

which one not working for which case can you say it clearly...?

Answer 12

import re
#your code goes here
num = input()

pathern = re.compile(r"^(1|8|9)(\d){7}quot;)

if pathern.search(num):
print("Valid")
else:
print("Invalid")

Answer 13

import re
#your code goes here
num=int(input())
pattern=r"[189][0-9][0-9][0-9][0-9][0-9][0-9][0-9]"
if re.match(pattern,"num"):
print('Valid')
else:
print('Invalid')
It's not working only 3/5

Answer 14

Building upon Prerna Vijay Lohar's answer:

import re

num = str(input())
pattern = "\A(1|8|9)\d{7}quot;

# \A - an anchor dictates a particular location in the search string where a match must occur.
# Anchor a match to the start of <string>.

# (1|8|9) - matches any 1, 8, or 9 at the start

# \d{7} - \d matches any digits subsequent, {7} matches seven times (total of 8 from above)

# $ indicates must end with the seven digits

if re.match(pattern,num):
print("Valid")
else:
print("Invalid")

Answer 15

import re

pattern = "[189]\d{7}quot;
num = input()

match = re.match(pattern,num)

if match:
print("Valid")
else:
print("Invalid")

Answer 16

import re
#your code goes here
num = str(input())
pattern = "\A(1|8|9)\d{7}quot;

if re.match(pattern,num):
print("Valid")
else:
print("Invalid")

I got my answer with this code

Answer 17

import re
num = input()

pathern = re.compile(r"^(1|8|9)(\d){7}quot;)

if pathern.search(num):
print("Valid")
else:
print("Invalid")

Answer 18

import re

num = str(input())
pattern = r"\A(1|8|9)\d{7}quot;

if re.match(pattern,num):
print("Valid")
else:
print("Invalid")

can't be better than this

Answer 19

import re
#your code goes here
num = str(input())
pattern = "\A(1/8/9)\d{7}quot;
if re.match(pattern,num):
print("Valid")
else:
print("Invalid")

Answer 20

import re

def valid(s):
pattern = re.compile("^[189]\d{7}quot;)
return pattern.match(s)

s = input("")
if (valid(s)):
print("Valid")
else:
print("Invalid")

THIS IS THE CORRECT ONE