Question

我正在尝试更换列表中的项目，而不是最后的项目，只是原始项目的位置，然后在要求您输入密码时需要新项目。我是Python的新手。

location ~ \.php$ { try_files $uri =404; fastcgi_split_path_info ^(.+\.php)(/.+)$; fastcgi_pass unix:/var/run/php5-fpm.sock; fastcgi_index index.php; fastcgi_param SCRIPT_FILENAME $document_root$fastcgi_script_name; include fastcgi_params; }

Answer 1

要替换列表中的项目，只需按列表位置分配：

mylist = ['hello', 'goodbye', 'goodnight']

mylist[2] = 'good day'

Answer 2

如果你知道列表中元素的价值（让我们说你要改变F Block的密码，'mrjoefblock'），你可以使用{{ 3}}函数返回索引，就像那样：

In [1]: passwords = ['mrjoebblock' , 'mrjoefblock' , 'mrjoegblock', 'mrjoeadmin' ]

In [2]: passwords.index('mrjoefblock')
Out[2]: 1

In [3]: passwords[passwords.index('mrjoefblock')] = 'newjoefblock'

In [4]: passwords
Out[4]: ['mrjoebblock', 'newjoefblock', 'mrjoegblock', 'mrjoeadmin']

通过这种方式，您可以更新密码并保持相同的位置（如果您要问的话）。

但是我改变这样的密码似乎很奇怪。我不是把它作为一个列表，而是改为字典：

In [1]: credentials = {
   ...:     'B Block': 'mrjoebblock',
   ...:     'F Block': 'mrjoefblock',
   ...:     'G Block': 'mrjoegblock',
   ...:     'Admin': 'mrjoeadmin'
   ...: }

我们现在可以找到具有给定密码的所有用户：

In [2]: [user for user, password in credentials.iteritems() if password == 'mrjoeadmin']
Out[2]: ['Admin']

并且可以轻松更改任何用户的密码：

In [3]: credentials['F Block'] = 'newjoefblock'

In [4]: credentials
Out[4]:
{'Admin': 'mrjoeadmin',
 'B Block': 'mrjoebblock',
 'F Block': 'newjoefblock',
 'G Block': 'mrjoegblock'}

对不起，最初还不清楚你在问什么。

我提出了这个简单的例子：

def login():
    """
    This function asks for the password and returns either the username (if logged in successfully)
    or False (if the password was incorrect).
    """
    password = raw_input("Password: ")

    if password == 'admin':
        return 'Administrator'
    elif password == 'ggguest':
        return 'Guest'
    else:
        return False


if __name__ == '__main__':
    # We'll introduce a new variable which will be either a logged in user's name
    # or False (based on login()'s function result)
    user_logged_in = login()

    while True:  # An endless loop
        if user_logged_in:
            # the login() function didn't return False, so we're logged in!
            print "Yay! You're now logged in as {0}!".format(user_logged_in)

            # Here we'll do whatever you need to. 
            # I kept the example simple so it just asks if the user wants to log out. 
            # You can change the user's password here and ask for the password again.

            want_to_logout = raw_input("Want to log out? [y/n]: ")

            if want_to_logout == 'y':
                user_logged_in = login()  # Ask for the password again
            else:
                print 'Alright then! Do your thing!'
                break  # Break out of loop and continue
        else:
            # wrong password, ask again
            print "Wrong password :("
            user_logged_in = login()  # Ask for the password again

    print "Keep up the good job, {0}!".format(user_logged_in)

以及它的工作方式：

$ python admin.py
Password: wrong_password
Wrong password :(
Password: admin
Yay! You're now logged in as Administrator!
Want to log out? [y/n]: y
Password: ggguest
Yay! You're now logged in as Guest!
Want to log out? [y/n]: n
Alright then! Do your thing!
Keep up the good job, Guest!

这样，您可以在需要时继续询问用户密码，否则继续。

尝试将列表中的项目替换为其他项目

2 个答案: