我是python的新手,我写了一个简单的登录脚本。 我不明白代码有什么问题: 1.)我使用过getpass.getpass(prompt =“输入密码:”)),但它在运行时显示密码,并显示警告。如何忽略它。
2.。)如何使用try-except来捕获错误的登录信息。
class login:
print "Login Page"
def __init__(self,id,pas):
self.id="admin"
self.pas="admin"
def check(self,id,pas):
if(self.id==log.id and self.pas==log.pas):
print "Login success!"
else:
print "who r u???"
log=login("","")
import getpass
log.check(raw_input("Enter Login ID:"),getpass.getpass(prompt="Enter password: "))
输出:
Login Page
Enter Login ID:ad
Warning (from warnings module):
File "C:\Python27\lib\getpass.py", line 92
return fallback_getpass(prompt, stream)
GetPassWarning: Can not control echo on the terminal.
Warning: Password input may be echoed.
Enter password: ad
who r u???
答案 0 :(得分:0)
在Windows上工作正常,试试这个:
class login:
print "Login Page"
def __init__(self, id, pas):
self.id = "admin"
self.pas = "admin"
def check(self, id, pas):
if(self.id == id and self.pas == pas):
print ("Login success!")
else:
print ("who r u???")
if __name__ == '__main__':
log = login("", "")
id1 = raw_input("Enter Login ID: ")
pas1 = raw_input("Enter password: ")
log.check(id1, pas1)