Question

此代码的前提是要求输入一个名称，最多3次。

password = 'correct'
attempts = 3
password = input ('Guess the password: ')
while password != 'correct' and attempts >= 2:
    input ('Try again: ')
    attempts = attempts-1
if password == 'correct':               #Where the problems begin
    print ('Well done')

我只能输入正确的密码，以便第一次返回“做得好”。＆＃39;在另外两次尝试中，它会再次尝试返回。＆＃39;如果在任何尝试中输入，我怎样才能让它完美地返回？

Answer 1

如果您想再试一次，那么您需要捕获该值。

password = input ('Try again: ')

否则，while循环永远不会停止。

此外，Python有while-else，可以帮助您调试问题

while password != 'correct' and attempts >= 2:
    password = input ('Try again: ')
    attempts = attempts-1
else:
    print('while loop done')
    if password == 'correct':               #Where the problems begin
        print ('Well done')

或

attempts = 3
password = input('Enter pass: ')
while attempts > 0:
    if password == 'correct':
        break
    password = input ('Try again: ')
    attempts = attempts-1
if attempts > 0 and password == 'correct':
    print ('Well done')

Answer 2

attempts = 3
while attempts > 1:
    password = input("Guess password")
    if password == "correct" and attempts > 2:
        print("Well done")
        break
    else:
        print("try again")
    attempts = attempts - 1

Answer 3

这是一种有趣的方式，而不是使用计数器，您可以创建一个包含三个元素的列表，$$watchers测试确保仍有剩余元素：

while

将if语句添加到while循环

3 个答案: