Question

以下是根据用户输入创建HTTP或FTP连接的代码。 if和elif条件总是以某种方式评估为FALSE。输入1和0两个打印＆＃39;抱歉，错误答案＆＃39;。

domain = 'ftp.freebsd.org'
path = '/pub/FreeBSD/'

protocol = input('Connecting to {}. Which Protocol to use? (0-http, 1-ftp): '.format(domain))
print(protocol)
input()

if protocol == 0:
    is_secure = bool(input('Should we use secure connection? (1-yes, 0-no): '))
    factory = HTTPFactory(is_secure)
elif protocol == 1:
    is_secure = False
    factory = FTPFactory(is_secure)
else:
    print('Sorry, wrong answer')
    import sys
    sys.exit(1)

connector = Connector(factory)

try:
    content = connector.read(domain, path)
except URLError as e:
    print('Can not access resource with this method')
else:
    print(connector.parse(content))

输出：

Connecting to ftp.freebsd.org. Which Protocol to use? (0-http, 1-ftp): 0
0

Sorry, wrong answer
$ python abstractfactory.py
Connecting to ftp.freebsd.org. Which Protocol to use? (0-http, 1-ftp): http
http

Sorry, wrong answer
$ python abstractfactory.py
Connecting to ftp.freebsd.org. Which Protocol to use? (0-http, 1-ftp): 1
1

Sorry, wrong answer

$

请指教。我在这做错了什么？感谢。

Answer 1

Python 3中的输入，它看起来像你正在使用，以字符串形式出现。您需要通过int()强制转换它（尽管这需要在输入错误的情况下谨慎处理并进行异常处理），以便将其与整数进行比较。

Answer 2

Python input（）将输入作为Unicode字符串，您需要显式地将输入作为整数与0比较，

if int(input) == 0:
    # Do something
elif int(input) == 1:
    # Do something

Python，如果条件始终评估为false，则无论用户输入如何

2 个答案: