How to exactly match string when there is variable in regex - python-3.6

Question

I'm Trying to create phone book program using python3. If name/phone number given already exist in phone book i need to inform the user to give unique name/number. Below is my code
import re,os
take_input='yes'
def duplicate_exists(arg):
if os.path.exists('ph_dir.txt'):
with open('ph_dir.txt','r') as rp:
contents=rp.read()
res=re.findall(arg,contents)
if res:
print(f"{arg} already exist.")
tk_ip='yes'
else:
tk_ip='no'
return tk_ip
while take_input == 'yes':
name=input("Enter the name:")
phno=input("Enter phone number:")
name=name.strip()
phno=phno.strip()
take_input=duplicate_exists(name)
take_input=duplicate_exists(phno)
with open('ph_dir.txt','a') as wp:
contents=wp.write(f'\n{name}:{phno}')
In the above code i'm using duplicate_exists() to find the duplicate for both name and phno
ph_dir.txt already has below entry
abcd:12345
now if i give the input as abc and 123 it says already exists.
Improper op:
Enter the name:abc
Enter phone number:123
abc already exist.
123 already exist.
Enter the name:
I tried word boundary res=re.findall(\barg\b,contents) in regex. But this will give syntax error SyntaxError: unexpected character after line continuation character. I'm using same function to find duplicate of phone number also. So i cant hard code any value here. How to solve this problem?

Answer 1

use this regex: re.findall(f'{arg}:.*|.*:{arg}', name)
def duplicate_exists(arg):
if os.path.exists('ph_dir.txt'):
with open('ph_dir.txt','r') as rp:
contents=rp.read()
res=re.findall(f'{arg}:.*|.*:{arg}', contents)
if res:
print(f"{arg} already exist.")
tk_ip='yes'
else:
tk_ip='no'
return tk_ip
also you can check both name and phno in one regex matching:
def duplicate_exists(name, phno):
if os.path.exists('ph_dir.txt'):
with open('ph_dir.txt','r') as rp:
contents=rp.read()
res=re.findall(f'{name}:.*|.*:{phno}', contents)
if res:
print(f"{name} or {phno} already exist.")
tk_ip='yes'
else:
tk_ip='no'
return tk_ip