Question

刚刚开始使用python并绞尽脑汁，但似乎无法做到这一点。

print('Enter correct username and password combo to continue')
count=0
password=Hytu76E
username=bank_admin

while password!='Hytu76E' and username!='bank_admin' and count<4:
    username=input('Enter username: ') and password=input('Enter password: ')

    if password=='Hytu76E' and username=='bank_admin':
     print('Access granted')

    else:
        print('Access denied. Try again.')
        count-=1

语法错误，无法在第6行上分配给操作员username = input。

Answer 1

这里试试这个（我尽量少改变你的代码，以便你自己识别相同的逻辑）

print('Enter correct username and password combo to continue')
count = 0

# "" or '' because you are assigning a value string into it
password = ""
username = ""

# looping will continue when wrong input for three times and ask again...
while password!='Hytu76E' and username!='bank_admin' and count < 3:
    # you are collecting user input from CLI separately (you can not assign and operator to such operation as per your code ;)
    username = input("Enter username: ")
    password = input("Enter password: ")

    if password=='Hytu76E' and username=='bank_admin':
     # if match, grand and break
     print('Access granted')
     break

    else:
        print('Access denied. Try again.')
        count+=1     # as per gbse, in the comments, you will need the + to count up

代码中的问题：

# you are assigning string value, what for? this would make the loop hit positive the first time
password=Hytu76E       # string assignment error in syntax, anyway
username=bank_admin    # string assignment error in syntax, anyway

# you can not assigning and operator in the input because of no if condition in this line, also you should compare the values of the input
username=input('Enter username: ') and password=input('Enter password: ')

# if code is ok, then move outside the loop in the case when the user enters the first time good answers
if password=='Hytu76E' and username=='bank_admin':
   print('Access granted')

    else:
        print('Access denied. Try again.')

        # you are decremented the counter which would never leave teh loop at 4, you should add one on each iteration so count+=1 (count = count + 1) 
        count-=1

Answer 2

修复了实现您要执行的操作的代码：

print('Enter correct username and password combo to continue')
count=0
while count < 3:
    username = input('Enter username: ')
    password = input('Enter password: ')
    if password=='Hytu76E' and username=='bank_admin':
        print('Access granted')
        break
    else:
        print('Access denied. Try again.')
        count += 1

已做出的更改：

删除了username和password的定义，因为它是多余的，可以省略
将while语句更改为计算count
仅在if语句中验证凭据，而不是在while
将count的减少量更改为增加（从count -=更改为count +=）
break输入正确凭据时的循环

Answer 3

我认为这就是您正在寻找的内容：接受用户名和密码，并根据代码中提到的特定内容进行验证，最大限制为3

print('Enter correct username and password combo to continue')
count=1

while count<4:
    username=input('Enter username: ')
    password=input('Enter password: ')
    if password=='Hytu76E' and username=='bank_admin':
        print('Access granted')
        count=5
    else:
        print('Access denied. Try again.')
        count+=1

Answer 4

首先，您可以删除在开始时为密码和用户名提供的初始定义，以及将while循环更改为while count＆lt; 4

所以它看起来像：

print('enter the correct username and password combo to continue')
count = 0
while count<4:

如果我们保留了它以前的状态，那就更不必要了，并使你的程序更加混乱。

要修复语法错误，您需要删除并放置在用户名和密码之间，所以中间看起来更像是这样：

username = input('Enter username: ')
password = input('Enter password: ')

然后在结束时你想要将count- = 1改为count + = 1，因为如果它每次都不会达到4而你的循环将是无限的，这不是你想要实现的。

以下是整个修复：

print('Enter correct username and password combo to continue')
count=0
while count<4:
    username=input('Enter username: ')
    password=input('Enter password: ')
    if password=='Hytu76E' and username=='bank_admin':
        print('Access granted')
        count=5
    else:
        print('Access denied. Try again.')
        count+=1

以下是我所做的更改列表：

删除了第3行和第4行中的密码和用户名定义
将while循环更改为while＆lt; 4
删除了＆inbetween username = input和password = input
在if语句后添加count = 5，以便循环结束

Python用户名和密码，共3次尝试

4 个答案: