所以我没有收到任何错误,但它没有正确运行。用# **
突出显示的代码位是出错的地方。因此,当我运行它时,无论是否放入.com或.co.uk或.org.uk,它总是打印出我的代码无效。
import time
#Imports the time library.
#3
#Email validator
#Make a program to check whether an email address is valid or not.
#You could make sure that there are no spaces, that there is an @ symbol and a dot somewhere after it. Also check that the end parts of the address are not blank.
#Extensions:
#1. When an email address is found to be invalid, tell the user exactly what they did wrong with their email address rather than just saying it is invalid
#2. Allow the user to choose to give a text file with a list of email addresses and have it process them all automatically.
print("Only .com or .co.uk or .org.uk are accepted.")
def ev():
#starts the definition and defines the command.
time.sleep(1)
#1 second wait
email = input("Email: ")
dot = "."
at = "@"
space = " "
com = ".com"
couk = ".co.uk"
org = ".org.uk"
if at not in email:
print("Invalid. There is no @ in your email.")
#Says email is invalid as there is no @
ev()
elif dot not in email:
print("Invalid. There is no .(dot) in your email.")
#Says email is invalid as there is no .
ev()
#Loops to asking for the email again
elif space in email:
print("Invalid. There shouldn't be any spaces in your email.")
#Says email is invalid as there is a space
ev()
#Loops to asking for the email again
elif com not in email or couk not in email or org not in email: # **
print("This email is not valid. Only .com or .co.uk or .org.uk are accepted.")**
ev()
#Loops to asking for the email again
else:
print("Valid!")
ev()
#Ends the definition so it starts automatically.
答案 0 :(得分:3)
您需要and
而不是or
:
试试这个:
elif com not in email and couk not in email and org not in email:
print("This email is not valid. Only .com or .co.uk or .org.uk are accepted.")**
ev()
#Loops to asking for the email again