python implements a simple login system

Novice, although relatively simple things, but made me for a long time.

Many imperfections, such as the lock user, the same user entered the wrong password three times will be locked, but if in the second and third user entered the original user, the count will be recalculated.

In addition does not exist the user input password error too many times will also be locked, and then can create a locked user, this will not be done, as long as the detection 1 whether there is a user name can be.

The solution is to add a special count variable in a special file and discard the file blocklist, but I don't know how to operate it, so I will keep it.

Although python does not have main functions like C1 as the program entry, let's make one, easy to understand

''' Log in the system 
 Requirements: 
1 You can choose to create a new user 
2 , can log in the original user 
3 Enter the wrong password 3 Then the user is locked '''
 
flag=True # Loop control character 
 
def createuser():
  f=open('userlist.txt','r')# Open the file for an existing user, assuming that the file already exists 
  flag=True
  name=f.readlines()
  f.close()
  while flag:
    username=input('username:')
    flag2=False# A marker that already exists for the user name 
    for line in name:
      if(username==line.split('*')[0]):
        flag2=True
        print(" The user name already exists, please re-enter it ")
    if flag2!=True:
      f=open('userlist.txt','a')# Create a new user 
      f.write('\n'+username)
      f.write('*')
      password=input('password:')
      f.write(password)
      f.close()
      break
  main()
 
 
def login():# Login function, enter password error 3 Then lock the user 
  count=0# Password error count ,3 Time is locked 
  f=open('userlist.txt','r')
  info=f.readlines()
  f.close()
  user=None# Repeat the user token 
  while flag:
    flag2=False
    f2=open('blocklist.txt', 'r')
    block_name=f2.readlines()
    f2.close()
    username=input('username:')
    if user==None:#user If it is not used, the user name entered is given directly 
      user=username
    elif user!=username:# If the 1 The user name entered for the second time is not 1 Sample, then on the record 1 The second user name, while counting to zero 
      user=username
      count=0;
    for line in block_name:# Check if the user name is locked, and lock returns to the main menu 
      if username==line.strip('\n'):
        print(' This user has been locked. Please contact your administrator ')
        main()
    password=input('password:')
    for line in info:
      if(username==line.split('*')[0]and password==line.split('*')[1].strip('\n')):
        print(' Login successful! ')
        flag2=True
    if flag2==False:
      count+=1
      print(" User name or password is incorrect, please re-enter ")
    if count==3:# error 3 Second, add the user name to the lock list 
      print(' Too many password errors, the user has been locked ')
      f3=open('blocklist.txt','a')
      f3.write('\n'+username)
      f3.close()
      count=0# Reset the count after joining the blacklist 
      main()# Add the blacklist and return to the main menu 
info='''
------ Please enter the relevant number -----
1. Create a new user 
2. Log in to existing users 
3. Exit the program 
'''
 
def main():
  print(info)
  while flag:
    i=input()
    if i=='1':
      createuser()
      break
    elif i=='2':
      login()
      break
    elif i=='3':
      exit()
    else:
      print(" Please enter the correct number .")
 
main()# Program entrance