使用 jack 登录工作正常，但即使认为 jake 存在于阵列中，我也无法使用 jake 登录。它只是跳过整个for循环“for i in storedusername”。如有帮助，将不胜感激。
代码如下：
import pygame
import sys
import random
storedusername = ["jack","jack","jack","jake","jack","jack","jack",] # The place where the username is stored
storedpass = ["abcde","abcde","abcde","12345","abcde","abcde","abcde",] # The current place where the password is stored
Login = False
def login(): # Function used to login
global Login # Global the logins fucntion
Login = False # Sets Login to false
user = input("Enter Username: ") # User enters the username that they would like to login with
print(user)
for i in storedusername: # Loops through the items in the list
if i == user: # Compares the items in the list with the username that the user has entered
print("user found") # If the user was found then the program will tell the user that
pos = int(storedusername.index(user)) # Finds the position where the username is stored
print(pos)
for j in range(0,10): # Has 10 tries to do this loop
password = input("Enter Password: ") # The user enters the password which they think matches with the username.
#for i in range(0,10):
if password == storedpass[pos]: # Goes to the position where the password matches the username is stored and compares the values.
print("password match username") # Program returns if the username and password match
Login = True # Turns login to True
return Login # Returnes Login
break
else:
print("Pasword does not match try again") # If the password does not match then the program will notify it.
print("too many attempts close the program") # If there are too many attempts than it will close.
else:
print("not found") # If the username is not found than it will be promted that it is not found.
return Login # Returns login.
login()
答案 0 :(得分：1)
您在外循环和内循环中都使用了变量 i，因此它在内部发生了变化。
必须是：
for i in ... :
...
for j in ... :
...
答案 1 :(得分：0)
您可能会看到由于缩进错误而无法正常工作
else:
print(
"Pasword does not match try again") # If the password does not match then the program will notify it.
print(
"too many attempts close the program") # If there are too many attempts than it will close.
阻止。
请把它移到左边，应该会更好。
修改后的版本
storedusername = ["jack", "jack", "jack", "jake", "jack", "jack",
"jack", ] # The place where the username is stored
storedpass = ["abcde", "abcde", "abcde", "12345", "abcde", "abcde",
"abcde", ] # The current place where the password is stored
(width, height) = (644, 412)
Login = False
def login(): # Function used to login
global Login # Global the logins fucntion
Login = False # Sets Login to false
user = input(
"Enter Username: ") # User enters the username that they would like to login with
print(user)
for i in storedusername: ############ It skips this loop
if i == user:
print("user found")
pos = int(storedusername.index(user))
print(pos) ########### All the way up to here
for i in range(0, 10): # Has 10 tries to do this loop
password = input(
"Enter Password: ") # The user enters the password which they think matches with the username.
# for i in range(0,10):
if password == storedpass[
pos]: # Goes to the position where the password matches the username is stored and compares the values.
print(
"password match username") # Program returns if the username and password match
Login = True # Turns login to True
return Login # Returnes Login
break
# --> wrong indentation
else:
print(
"Pasword does not match try again") # If the password does not match then the program will notify it.
print(
"too many attempts close the program") # If there are too many attempts than it will close.
# <-- wrong indentation
else:
print(
"not found") # If the username is not found than it will be promted that it is not found.
return Login # Returns login.
# register()
login()
允许
Enter Username: jack
jack
user found
0
Enter Password: abde
Pasword does not match try again
Enter Password: abcde
password match username
Process finished with exit code 0
我不打算修改您的解决方案的流程，但我建议在行之前而不是在其中写注释。此外，您应该避免变量阴影（在不同范围内使用相同的变量，如 i）。最后一件事是您在检查密码时缩进略有错误 - 即使没有匹配的名称，它也会运行。我允许自己提出所有这些修改的版本。请看一下：
def login():
""" Function used to login"""
# Global the logins fucntion
global Login
Login = False
# User enters the username that they would like to login with
user = input("Enter Username: ")
print(user)
for stored_user_name in stored_user_names:
if stored_user_name == user:
print("user found")
pos = int(stored_user_names.index(user))
# Has 10 tries to do this loop
for try_attempt in range(0, 10):
# The user enters the password which they think matches with the username.
password = input("Enter Password: ")
# Goes to the position where the password matches the username is stored and compares the values.
if password == storedpasswords[pos]:
# Program returns if the username and password match
print("password match username")
Login = True # Turns login to True
return Login # Returnss Login
else:
# If the password does not match then the program will notify it.
print("Password does not match try again")
# If there are too many attempts than it will close.
print("too many attempts close the program")
else:
# If the username is not found than it will be promted that it is not found.
print("not found")
return Login
答案 2 :(得分：0)
您认为缩进不正确是正确的。与之前一样，如果您输入的用户名不在列表中，那么 storedusername 中的每个用户仍会被要求输入密码十次（总共 70 次！）。将块从 for i in range(0,10) 移动一个缩进进一步修复它。
for i in storedusername: ############ It skips this loop
if i == user:
print("user found")
pos = int(storedusername.index(user))
print(pos) ########### All the way up to here
# >>>> indent here
for i in range(0,10): # Has 10 tries to do this loop
password = input("Enter Password: ") # The user enters the password which they think matches with the username.
#for i in range(0,10):
if password == storedpass[pos]: # Goes to the position where the password matches the username is stored and compares the values.
print("password match username") # Program returns if the username and password match
Login = True # Turns login to True
return Login # Returnes Login
break
else:
print("Pasword does not match try again") # If the password does not match then the program will notify it.
# >>>>>
print("too many attempts close the program") # If there are too many attempts than it will close.
else:
print("not found") # If the username is not found than it will be promted that it is not found.