Python 从用户输入的内容中读取下一行

Question

所以基本上我有一个程序，它可以在文档中创建用户名和密码，并搜索与使用该程序的人输入的用户名一起使用的密码。

例如： 程序要求我输入我输入的用户名：'13'。 文本文档中的 13 以下是“sadfsdfsdf”。 我希望程序跳到“用户名：1​​3”下方并读取并打印“密码：sadfsdfsdf”。

请注意，我在 .txt 文件中有多个用户名和密码

u = '用户名：'

提前致谢！

def SearchRecordProgram():
    while True:
                    user_search = input("Please enter the username you wish to see the password for: ")
                    full_user = u + user_search
                    if full_user in open('User_Details.txt').read():
                            print(" ")
                            print ("Username found in our records")
                            print(" ")
                            print(full_user)
                            break
                    else:
                            print("Username entered does not match our records")

Answer 1

所以，想象你的文件是这样的：

Username : 13
Password : sadfsdfsdf
Username : 15
Password : Lalalala

你可以像这样解析它（使用正则表达式）：

import re  #  regular expression module
regex = re.compile("Username :(.*)\nPassword :(.*)")

# read text from file
with open(filePath,'r') as f:
    text = f.read() 

u_pass = regex.findall(text)
# [(' 13', ' sadfsdfsdf'), (' 15', ' Lalalala')]

user_dict = {u:password for u,password in u_pass}
#  {' 13': ' sadfsdfsdf', ' 15': ' Lalalala'}

现在您可以通过询问该用户的密码来获取该用户的密码：

# given username and password_input

if username in user_dict and user_dict[username] == password_input:
# if username exists and password okay
    print "Authentication succeeded."

Answer 2

当你想打开一个文件时，你几乎应该总是使用with 语句：

with open('User_Details.txt') as read_file:
    # Do reading etc. with `read_file` variable

这将确保正确处理任何错误并且文件不会保持打开状态。

现在文件已打开，我们需要遍历每一行，直到找到与我们的用户名匹配的行。我希望你知道for loop 的工作原理：

username = 'Username: 13'  # Get this however you want
with open('User_Details.txt') as read_file:
    for line in read_file:
        line = line.strip()  # Removes any unnecessary whitespace characters
        if line == username:
            # We found the user! Next line is password

我们需要获取包含密码的下一行。有很多方法可以获取下一行，但一种简单的方法是使用 next() 函数，它简单地从可迭代对象中获取下一个元素（在本例中为文件中的下一行）：

username = 'Username: 13'
with open('User_Details.txt') as read_file:
    for line in read_file:
        line = line.strip()
        if line == username:
            password = next(read_file)
            break  # Ends the for loop, no need to go through more lines

现在您有了密码和用户名，您可以对它们做任何您想做的事情。将输入和输出置于程序逻辑之外通常是个好主意，因此不要直接在 for 循环内打印密码，而是直接在此处接收密码，然后在外部打印。

你甚至可能想把整个搜索逻辑变成一个函数：

def find_next_line(file_handle, line):
    """Finds the next line from a file."""
    for l in file_handle:
        l = l.strip()
        if l == line:
            return next(file_handle)


def main():
    username = input("Please enter the username you wish to see the password for: ")
    username = 'Username: ' + username
    with open('User_Details.txt') as read_file:
        password = find_next_line(read_file, username)
    password = password[len('Password: '):]
    print("Password '{0}' found for username '{1}'".format(password, username))

if __name__ == '__main__':
    main()

最后，以这种格式存储任何东西绝对是疯狂的（更不用说密码安全的东西，但我知道你只是在学习东西），为什么不这样做：

username:password
markus:MarkusIsHappy123
BOB:BOB'S PASSWORD

这可以很容易地转换成字典：

with open('User_Details.txt') as read_file:
    user_details = dict(line.strip().split(':') for line in read_file)

现在要获取用户名的密码，您可以：

username = input('Username: ')
if username in user_details:
    print('Password:', user_details[username])
else:
    print('Unknown user')

Answer 3

也许不是美女，但这样的事情会起作用：

with open(users_file, "r") as f:
    lines = f.read()

def get_password(user_id):
    entries = iter(lines.splitlines())
    for entry in entries:
        if(entry.startswith("{}:{}".format(prefix, user_id))):
            return next(entries)

print "Password:", get_password("13")

Answer 4

def SearchRecordProgram():
    while True:
        user_search = input("Please enter the username > ")

        file = open('User_Details.txt')
        usernames = file.readlines()  # Read file into list
        file.close()

        if user_search in usernames: # Username exists
            print ('Found Username')
            password = usernames[usernames.index(user_search)+1]
            print ('Password is: ' + password)   

            break  # Exit loop

         else:  # Username doesn't exist
             print("Username entered does not match our records")

请注意，如果密码恰好是用户名，这将不起作用，例如：

user1
password1
user2
user3  # This is the password for user3
user3
password3

如果您搜索“user3”，此代码将输出“user3”作为“user3”的密码，因为它会找到“user3”的第一个实例（第 4 行），然后查找下一行（第 5 行），这是下一个用户名。

更糟糕的是，如果“user3”是文件中的最后一行，它会以错误结束，因为没有更多的行。您可以使用以下代码添加检查找到的用户名的索引是否为偶数（即索引 0、2、4、8）：

if not usernames.index(user_search) % 2:  # Checks the remainder after / by 2.
                                          # If it's 0 then it's an even index.
    password = usernames[usernames.index(user_search)+1]
    print ('Password is: ' + password)   

但如果发生这种情况，您无能为力。

但是，您可以将用户名文件操作为只有这样的用户名：

lst = [0, 1, 2, 3, 4, 5, 6, 7]
print (lst[0::2])

打印出来的

[0, 2, 4, 6]

所以代码可以改成这样：

def SearchRecordProgram():
    while True:
        user_search = input("Please enter the username > ")

        usernames = open('User_Details.txt').readlines()  # Read file into list

        if user_search in usernames[0::2]: # Username exists
            print ('Found Username')
            password = usernames[1::2][usernames[0::2]].index(user_search)]
            print ('Password is: ' + password)   

            break  # Exit loop

         else:  # Username doesn't exist
             print("Username entered does not match our records")

这里是一个细分

password = usernames[1::2]  # Gets the odd items (passwords)
                          [  # Get index item. The index returned from searching
                             # the usernames will be the same as the index needed
                             # for the password, as looking at just the odd items:
                             # [0, 2, 4, 6]
                             # [1, 3, 5, 7] - even has the same index as next odd.
                           usernames[0::2] # Gets even items (usernames)
                                          ].index(user_search)]
                                          # Only searches the usernames. This
                                          # means passwords being the same is not
                                          # an issue - it doesn't search them