blocks|key|1268578|text|您可以调用login()函数两次。只有第二个调用使用返回值，所以您可以删除第一个调用|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1268579|def+main():
++++login()+++++++++#+<---+Remove+this+one
++++log+=+login()|code-block|syntax|javascript|1268580|1268581|您可能会考虑使用getpass。在您的password字段中使用此选项而不是raw_input将阻止密码显示在屏幕上。|1268582|import+getpass

...

def+login():
++++...
++++passw+=+getpass.getpass("Password:+")|1268583|然后，其余代码的行为相同，但在控制台上，输出如下所示：|1268584|Username:+max2
Password:
Login+successful!|1268585|请注意，尽管输入了有效的密码，"+password+:“行仍为空|1268586|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/2/library/getpass.html^0|5|7|0|0|0|8|7|12|9|8|7|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|13|8|@$9|14|A|15|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|17|8|@]|D|@]|E|$]]|$1|L|3|M|5|6|7|18|8|@$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]]|D|@$9|1D|A|1E|1|1F]]|E|$]]|$1|N|3|O|5|H|7|1G|8|@]|D|@]|E|$I|J]]|$1|P|3|Q|5|6|7|1H|8|@]|D|@]|E|$]]|$1|R|3|S|5|H|7|1I|8|@]|D|@]|E|$I|J]]|$1|T|3|U|5|6|7|1J|8|@]|D|@]|E|$]]|$1|V|3|-4|5|6|7|1K|8|@]|D|@]|E|$]]]|W|$X|$5|Y|Z|10|E|$11|12]]]]

You call your <code>login()</code> function two times. Only the second one utilizes the return value, so you can remove the first call

<pre><code>def main():
 login() # &lt;--- Remove this one
 log = login()
</code></pre>

<hr>

Something you may wish to consider using is <a href="https://docs.python.org/2/library/getpass.html" rel="nofollow"><code>getpass</code></a>. Using this instead of <code>raw_input</code> on your password field will prevent the password from being displayed on screen.

<pre><code>import getpass

...

def login():
 ...
 passw = getpass.getpass("Password: ")
</code></pre>

The rest of the code then behaves the same, but on the console, the output looks like this:

<pre><code>Username: max2
Password:
Login successful!
</code></pre>

Notice that the "Password:" line is empty, despite typing in a valid password

blocks|key|1268611|text|这里的问题是，您在main中调用了login()两次。要修复您的错误，只需删除第一个调用，因为返回甚至都没有使用。另外，我不确定这是否是故意的，但是：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1268612|if+(user+in+us)+and+(passw+in+pw):|code-block|syntax|javascript|1268613|将匹配('user'，'pass')，即使(us，pw)是('username'，'password')。改用==运算符。|1268614|至少有一件事，你应该考虑改变：|1268615|us,+pw+=+line.strip().split("%7C")|1268616|至：|1268617|us,+pw+=+line.strip().split("%7C",+1)|1268618|它只将该行拆分一次，否则密码将无法包含%7C|1268619|entityMap^0|9|4|H|7|0|0|1J|2|0|0|0|0|0|J|1|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|14|8|@$9|15|A|16|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|17|8|@]|D|@]|E|$]]|$1|O|3|P|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|Q|3|R|5|6|7|19|8|@]|D|@]|E|$]]|$1|S|3|T|5|H|7|1A|8|@]|D|@]|E|$I|J]]|$1|U|3|V|5|6|7|1B|8|@$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|W|3|-4|5|6|7|1E|8|@]|D|@]|E|$]]]|X|$]]

The problem here is that you are calling <code>login()</code> twice whitin <code>main</code>. To fix your bug, just remove the first call since the return is not even used. Also, I'm not sure whether it's intended but:

<pre><code>if (user in us) and (passw in pw):
</code></pre>

will match ('user', 'pass') even if (us, pw) is ('username', 'password'). Use <code>==</code> operator instead.

One least thing, you should consider changing:

<pre><code>us, pw = line.strip().split("|")
</code></pre>

To:

<pre><code>us, pw = line.strip().split("|", 1)
</code></pre>

Which splits the line only once, otherwise the passwords won't be able to contain <code>|</code>

blocks|key|1268631|text|问题是调用了两个login()实例。以下代码将重复登录，直到正确为止：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1268632|def+login():++++
++++user+=+raw_input("Username:+")
++++passw+=+raw_input("Password:+")
++++f+=+open("users.txt",+"r")
++++for+line+in+f.readlines():
++++++++us,+pw+=+line.strip().split("%7C")
++++++++if+(user+in+us)+and+(passw+in+pw):
++++++++++++print+"Login+successful!"
++++++++++++return+True
++++print+"Wrong+username/password"
++++return+False

def+menu():
++++#here's+a+menu+that+the+user+can+access+if+he+logged+in.

def+main():
++++global+True
++++while+True:
++++++++True+=+True
++++++++log+=+login()
++++++++if+log+==+True:
+++++++++++++menu()
+++++++++++++True+=+False


main()|code-block|syntax|javascript|1268633|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

The issue was that there were two instances of login() being called.
The following Code Will Repeat The Login until it is correct:

<pre><code>def login(): 
 user = raw_input("Username: ")
 passw = raw_input("Password: ")
 f = open("users.txt", "r")
 for line in f.readlines():
 us, pw = line.strip().split("|")
 if (user in us) and (passw in pw):
 print "Login successful!"
 return True
 print "Wrong username/password"
 return False

def menu():
 #here's a menu that the user can access if he logged in.

def main():
 global True
 while True:
 True = True
 log = login()
 if log == True:
 menu()
 True = False


main()
</code></pre>

<pre><code>def login(): 
 user = raw_input("Username: ")
 passw = raw_input("Password: ")
 f = open("users.txt", "r")
 for line in f.readlines():
 us, pw = line.strip().split("|")
 if (user in us) and (passw in pw):
 print "Login successful!"
 return True
 print "Wrong username/password"
 return False

def menu():
 #here's a menu that the user can access if he logged in.

def main():
 login()
 log = login()
 if log == True:
 menu()
</code></pre>

And the users.txt is:

<pre><code>john22|1234
max2|2211
jack6|1551
</code></pre>

The problem is, when I run the program and enter the correct Username and Password, it prints out "Login successful!" but it doesn't continue to menu(). Instead, it shows "Username: " and "Password: " again. However, when I enter the correct username and password again (same or different pair from users.txt), it enters the menu() and everything works fine. 
So my guess is that you have to, for some reason, enter the correct info twice, and I don't want that.
I searched everywhere but I can't figure out why is this happening...

Simple login function in Python

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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def login():        user = raw_input("Username: ")    passw = raw_input("Password: ")    f = open("users.txt", "r")    for line in f.readlines():        us, pw ...

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