Python :不可排序类型: NoneType() < NoneType()

Question

使用IPtools package，我试图查看ip地址是否在特定的范围内。这是我的代码：

for line in g:
    org= line.split("|")[0]
    ranges = ast.literal_eval(line.split('|')[1])
    for range in ranges:
        start,end = range
        range_s = IpRange(start,end)
        if '198.0.184.126' in range_s is True:
        print (range_s)

下面是我的文件的样子：

FOP Michail Mandryk |[('91.195.172.0', '91.195.173.255'), ('195.95.222.0', '195.95.223.255')]
Circle 3 Inc.|[('66.64.129.28', '66.64.129.31'), ('216.23.64.120', '216.23.64.120'), ('216.215.250.46', '216.215.250.47')]
a1web.com.br|[('50.116.92.89', '50.116.92.89')]
Shandong Dezhou decheng district government|[('61.133.124.64', '61.133.124.79')]
Global ICT Solutions (ShangHai) CO.,LTD|[('43.247.100.0', '43.247.103.255')]
VendorCert|[('173.1.96.112', '173.1.96.127')]
Lowell City Library|[('198.0.184.112', '198.0.184.127')]
abc|[('123.0.0.0/8' , '12.12.3.0/8')]

我得到了这个错误，但我找不到原因。有人能帮忙吗？

TypeError                                 Traceback (most recent call last)
<ipython-input-59-420def563a4e> in <module>()
     19 
     20 #         print (start,end)
---> 21         range_s = IpRange(start,end)
     22 #         if '198.0.184.126' in range_s is True:
     23         print (range_s)

/opt/miniconda3/lib/python3.4/site-packages/iptools/__init__.py in __init__(self, start, end)
    158         start = _address2long(start)
    159         end = _address2long(end)
--> 160         self.startIp = min(start, end)
    161         self.endIp = max(start, end)
    162         self._len = self.endIp - self.startIp + 1

TypeError: unorderable types: NoneType() < NoneType()

Answer 1

如果函数不能解析有效的IPv4或IPv6地址，则返回IPv6。

您的start地址和end地址都无法进行解析，而两个None值上的min()函数也会失败，但会出现异常：

>>> min(None, None)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unorderable types: NoneType() < NoneType()

三次检查您传入的地址，以确保它们是有效的。您可以捕获此异常以打印导致问题的值，例如：

try:
    range_s = IpRange(start,end)
except TypeError:
    print('Problematic addresses:', start, end)
    raise

另外，您不需要在is True语句中测试if。这就是if的作用所在。相比之下，链接语句并不意味着你认为它在任何情况下都意味着什么。严格使用：

if '198.0.184.126' in range_s:
    print (range_s)

当使用is True时，您实际上是在测试('198.0.184.126' in range_s) and (range_s is True)，这永远不会是真的。