<p>您不需要循环查看<code>AccountU</code>中的每个条目—您知道要查找哪个用户名,只需要相应的密码。如果使用<code>.index()</code>方法,可以在<code>AccountU</code>列表中找到用户名的索引。然后,假设<code>AccountP</code>与<code>AccountU</code>直接匹配,您可以在同一索引中找到正确的密码,如下所示:</p>
<pre><code>AccountU = ['Cam', 'Copper']
AccountP = ['Pop1234', 'What?']
def LoginA():
print ("=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=")
print ("A few clicks away from your selection!")
UserE = input("Username: ")
if UserE in AccountU:
PasswordF = AccountP[AccountU.index(UserE)]
PassE = input("Password: ")
if PasswordF == PassE:
Menu()
else:
print ("Incorrect Password")
else:
print ("Your account is not working.")
LoginA()
</code></pre>
<p>不过,最好使用字典来存储用户名和密码。这将确保正确的用户与正确的密码匹配。下面是一个如何工作的例子:</p>
<pre><code>accounts = {'Cam': 'Pop1234',
'Copper': 'What?'}
def LoginA():
print ("=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=")
print ("A few clicks away from your selection!")
UserE = input("Username: ")
if UserE in accounts.keys():
PassE = input("Password: ")
if accounts[UserE] == PassE:
Menu()
else:
print ("Incorrect Password")
else:
print ("Your account is not working.")
LoginA()
</code></pre>