Youtube link finder problem

Question

I DONT KNOW WHAT IS WRONG IN MY CODE, CAN HELP ME?

THIS IS MY CODE;

takeLink =list(input())
stri="".join(takeLink )

backSlash=0

if re.search("com",stri):
for i in range(len(takeLink)):
if takeLink[i]==r'=':
break
print("".join(takeLink[i+1:]))

else:
for i in range(len(takeLink
if takeLink[i]==r'/':
backSlash +=1
if backSlash == 3:
break
print ("".join(takeLink[i+1:]))

Answer 1

#The simplest answer😁

link =input('')
if '=' in link :
lf1 = link.split('=')
d = lf1[-1]
print(d)
else:
lf2= link.split('/')
m = lf2[-1]
print(m)

Answer 2

Your code was pulling a syntax error because "for i in range(len(takeLink" is missing ")):" to end it

takeLink =list(input())
stri="".join(takeLink )

backSlash=0

if re.search("com",stri):
for i in range(len(takeLink)):
if takeLink[i]==r'=':
break
print("".join(takeLink[i+1:]))

else:
for i in range(len(takeLink)):
if takeLink[i]==r'/':
backSlash +=1
if backSlash == 3:
break
print ("".join(takeLink[i+1:]))

Answer 3

like this?

takeLink =list(input())
stri="".join(takeLink )

backSlash=0

if re.search("com",stri):
for i in range(len(takeLink)):
if takeLink[i]==r'=':
break
print("".join(takeLink[i+1:]))

else:
for i in range(len(takeLink)):
if takeLink[i]==r'/':
backSlash +=1
if backSlash == 3:
break
print ("".join(takeLink[i+1:]))

if so, it did not work

Answer 4

Need to import re to use re.search also, my bad. This time I tested it directly in code coach and it works great.

import re

takeLink =list(input())
stri="".join(takeLink )

backSlash=0

if re.search("com",stri):
for i in range(len(takeLink)):
if takeLink[i]==r'=':
break
print("".join(takeLink[i+1:]))

else:
for i in range(len(takeLink)):
if takeLink[i]==r'/':
backSlash +=1
if backSlash == 3:
break
print ("".join(takeLink[i+1:]))