验证用户输入的IP有效的最佳方法是什么？它以字符串形式出现。
答案 0 :(得分：170)
不要解析它。请问。
import socket
try:
socket.inet_aton(addr)
# legal
except socket.error:
# Not legal
答案 1 :(得分：62)
import socket
def is_valid_ipv4_address(address):
try:
socket.inet_pton(socket.AF_INET, address)
except AttributeError: # no inet_pton here, sorry
try:
socket.inet_aton(address)
except socket.error:
return False
return address.count('.') == 3
except socket.error: # not a valid address
return False
return True
def is_valid_ipv6_address(address):
try:
socket.inet_pton(socket.AF_INET6, address)
except socket.error: # not a valid address
return False
return True
答案 2 :(得分：59)
IPy module（用于处理IP地址的模块）将为无效地址抛出ValueError异常。
>>> from IPy import IP
>>> IP('127.0.0.1')
IP('127.0.0.1')
>>> IP('277.0.0.1')
Traceback (most recent call last):
...
ValueError: '277.0.0.1': single byte must be 0 <= byte < 256
>>> IP('foobar')
Traceback (most recent call last):
...
ValueError: invalid literal for long() with base 10: 'foobar'
然而，就像Dustin的回答一样，它会接受像“4”和“192.168”之类的东西，因为如上所述，这些是IP地址的有效表示。
如果您使用的是Python 3.3或更高版本，它现在包含ipaddress module：
>>> import ipaddress
>>> ipaddress.ip_address('127.0.0.1')
IPv4Address('127.0.0.1')
>>> ipaddress.ip_address('277.0.0.1')
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/usr/lib/python3.3/ipaddress.py", line 54, in ip_address
address)
ValueError: '277.0.0.1' does not appear to be an IPv4 or IPv6 address
>>> ipaddress.ip_address('foobar')
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/usr/lib/python3.3/ipaddress.py", line 54, in ip_address
address)
ValueError: 'foobar' does not appear to be an IPv4 or IPv6 address
对于Python 2，如果安装python-ipaddress，则可以使用ipaddress获得相同的功能：
pip install ipaddress
此模块与Python 2兼容，并提供与Python 3.3之后的Python标准库中包含的ipaddress模块非常相似的API。更多细节here。在Python 2中，您需要将IP地址字符串显式转换为unicode：ipaddress.ip_address(u'127.0.0.1')。
答案 3 :(得分：44)
从Python 3.4开始，检查IPv6或IPv4地址是否正确的最佳方法是使用Python标准库模块ipaddress - IPv4 / IPv6操作库s.a. https://docs.python.org/3/library/ipaddress.html获取完整的文档。
示例：
#!/usr/bin/env python
import ipaddress
import sys
try:
ip = ipaddress.ip_address(sys.argv[1])
print('%s is a correct IP%s address.' % (ip, ip.version))
except ValueError:
print('address/netmask is invalid: %s' % sys.argv[1])
except:
print('Usage : %s ip' % sys.argv[0])
对于其他版本：Github，phihag / Philipp Hagemeister，“Python 3.3的旧版Python版ipaddress”，https://github.com/phihag/ipaddress
来自phihag的后端可用，例如在Anaconda Python 2.7＆amp;安装程序中包含。股份公司https://docs.continuum.io/anaconda/pkg-docs
使用pip进行安装：
pip install ipaddress
s.a。：ipaddress 1.0.17，“IPv4 / IPv6操作库”，“3.3+ ipaddress模块的端口”，https://pypi.python.org/pypi/ipaddress/1.0.17
答案 4 :(得分：43)
def is_valid_ip(ip):
"""Validates IP addresses.
"""
return is_valid_ipv4(ip) or is_valid_ipv6(ip)
的IPv4：
def is_valid_ipv4(ip):
"""Validates IPv4 addresses.
"""
pattern = re.compile(r"""
^
(?:
# Dotted variants:
(?:
# Decimal 1-255 (no leading 0's)
[3-9]\d?|2(?:5[0-5]|[0-4]?\d)?|1\d{0,2}
|
0x0*[0-9a-f]{1,2} # Hexadecimal 0x0 - 0xFF (possible leading 0's)
|
0+[1-3]?[0-7]{0,2} # Octal 0 - 0377 (possible leading 0's)
)
(?: # Repeat 0-3 times, separated by a dot
\.
(?:
[3-9]\d?|2(?:5[0-5]|[0-4]?\d)?|1\d{0,2}
|
0x0*[0-9a-f]{1,2}
|
0+[1-3]?[0-7]{0,2}
)
){0,3}
|
0x0*[0-9a-f]{1,8} # Hexadecimal notation, 0x0 - 0xffffffff
|
0+[0-3]?[0-7]{0,10} # Octal notation, 0 - 037777777777
|
# Decimal notation, 1-4294967295:
429496729[0-5]|42949672[0-8]\d|4294967[01]\d\d|429496[0-6]\d{3}|
42949[0-5]\d{4}|4294[0-8]\d{5}|429[0-3]\d{6}|42[0-8]\d{7}|
4[01]\d{8}|[1-3]\d{0,9}|[4-9]\d{0,8}
)
$
""", re.VERBOSE | re.IGNORECASE)
return pattern.match(ip) is not None
的IPv6：
def is_valid_ipv6(ip):
"""Validates IPv6 addresses.
"""
pattern = re.compile(r"""
^
\s* # Leading whitespace
(?!.*::.*::) # Only a single whildcard allowed
(?:(?!:)|:(?=:)) # Colon iff it would be part of a wildcard
(?: # Repeat 6 times:
[0-9a-f]{0,4} # A group of at most four hexadecimal digits
(?:(?<=::)|(?<!::):) # Colon unless preceeded by wildcard
){6} #
(?: # Either
[0-9a-f]{0,4} # Another group
(?:(?<=::)|(?<!::):) # Colon unless preceeded by wildcard
[0-9a-f]{0,4} # Last group
(?: (?<=::) # Colon iff preceeded by exacly one colon
| (?<!:) #
| (?<=:) (?<!::) : #
) # OR
| # A v4 address with NO leading zeros
(?:25[0-4]|2[0-4]\d|1\d\d|[1-9]?\d)
(?: \.
(?:25[0-4]|2[0-4]\d|1\d\d|[1-9]?\d)
){3}
)
\s* # Trailing whitespace
$
""", re.VERBOSE | re.IGNORECASE | re.DOTALL)
return pattern.match(ip) is not None
IPv6版本使用“(?:(?<=::)|(?<!::):)”，在支持具有环视功能的条件的正则表达式引擎上可以替换为“(?(?<!::):)”。 （即PCRE，.NET）
修改
<强> EDIT2：
我找到了一些讨论如何用正则表达式解析IPv6地址的链接：
<强> EDIT3：
最后设法编写了一个通过所有测试的模式，并且我也很满意。
答案 5 :(得分：12)
我希望它足够简单和pythonic：
def is_valid_ip(ip):
m = re.match(r"^(\d{1,3})\.(\d{1,3})\.(\d{1,3})\.(\d{1,3})$", ip)
return bool(m) and all(map(lambda n: 0 <= int(n) <= 255, m.groups()))
答案 6 :(得分：7)
我必须为Markus Jarderot的帖子提供大量信誉 - 我的大多数帖子都是从他的文章中获得灵感。
我发现Markus的回答仍然没有通过他的回答引用的Perl脚本中的一些IPv6示例。
这是我的正则表达式，它传递了Perl脚本中的所有示例：
r"""^
\s* # Leading whitespace
# Zero-width lookaheads to reject too many quartets
(?:
# 6 quartets, ending IPv4 address; no wildcards
(?:[0-9a-f]{1,4}(?::(?!:))){6}
(?:25[0-4]|2[0-4]\d|1\d\d|[1-9]\d|\d)
(?:\.(?:25[0-4]|2[0-4]\d|1\d\d|[1-9]\d|\d)){3}
|
# 0-5 quartets, wildcard, ending IPv4 address
(?:(?:[0-9a-f]{1,4}(?::(?!:))){0,4}[0-9a-f]{1,4})?
(?:::(?!:))
(?:25[0-4]|2[0-4]\d|1\d\d|[1-9]\d|\d)
(?:\.(?:25[0-4]|2[0-4]\d|1\d\d|[1-9]\d|\d)){3}
|
# 0-4 quartets, wildcard, 0-1 quartets, ending IPv4 address
(?:(?:[0-9a-f]{1,4}(?::(?!:))){0,3}[0-9a-f]{1,4})?
(?:::(?!:))
(?:[0-9a-f]{1,4}(?::(?!:)))?
(?:25[0-4]|2[0-4]\d|1\d\d|[1-9]\d|\d)
(?:\.(?:25[0-4]|2[0-4]\d|1\d\d|[1-9]\d|\d)){3}
|
# 0-3 quartets, wildcard, 0-2 quartets, ending IPv4 address
(?:(?:[0-9a-f]{1,4}(?::(?!:))){0,2}[0-9a-f]{1,4})?
(?:::(?!:))
(?:[0-9a-f]{1,4}(?::(?!:))){0,2}
(?:25[0-4]|2[0-4]\d|1\d\d|[1-9]\d|\d)
(?:\.(?:25[0-4]|2[0-4]\d|1\d\d|[1-9]\d|\d)){3}
|
# 0-2 quartets, wildcard, 0-3 quartets, ending IPv4 address
(?:(?:[0-9a-f]{1,4}(?::(?!:))){0,1}[0-9a-f]{1,4})?
(?:::(?!:))
(?:[0-9a-f]{1,4}(?::(?!:))){0,3}
(?:25[0-4]|2[0-4]\d|1\d\d|[1-9]\d|\d)
(?:\.(?:25[0-4]|2[0-4]\d|1\d\d|[1-9]\d|\d)){3}
|
# 0-1 quartets, wildcard, 0-4 quartets, ending IPv4 address
(?:[0-9a-f]{1,4}){0,1}
(?:::(?!:))
(?:[0-9a-f]{1,4}(?::(?!:))){0,4}
(?:25[0-4]|2[0-4]\d|1\d\d|[1-9]\d|\d)
(?:\.(?:25[0-4]|2[0-4]\d|1\d\d|[1-9]\d|\d)){3}
|
# wildcard, 0-5 quartets, ending IPv4 address
(?:::(?!:))
(?:[0-9a-f]{1,4}(?::(?!:))){0,5}
(?:25[0-4]|2[0-4]\d|1\d\d|[1-9]\d|\d)
(?:\.(?:25[0-4]|2[0-4]\d|1\d\d|[1-9]\d|\d)){3}
|
# 8 quartets; no wildcards
(?:[0-9a-f]{1,4}(?::(?!:))){7}[0-9a-f]{1,4}
|
# 0-7 quartets, wildcard
(?:(?:[0-9a-f]{1,4}(?::(?!:))){0,6}[0-9a-f]{1,4})?
(?:::(?!:))
|
# 0-6 quartets, wildcard, 0-1 quartets
(?:(?:[0-9a-f]{1,4}(?::(?!:))){0,5}[0-9a-f]{1,4})?
(?:::(?!:))
(?:[0-9a-f]{1,4})?
|
# 0-5 quartets, wildcard, 0-2 quartets
(?:(?:[0-9a-f]{1,4}(?::(?!:))){0,4}[0-9a-f]{1,4})?
(?:::(?!:))
(?:(?:[0-9a-f]{1,4}(?::(?!:))){0,1}[0-9a-f]{1,4})?
|
# 0-4 quartets, wildcard, 0-3 quartets
(?:(?:[0-9a-f]{1,4}(?::(?!:))){0,3}[0-9a-f]{1,4})?
(?:::(?!:))
(?:(?:[0-9a-f]{1,4}(?::(?!:))){0,2}[0-9a-f]{1,4})?
|
# 0-3 quartets, wildcard, 0-4 quartets
(?:(?:[0-9a-f]{1,4}(?::(?!:))){0,2}[0-9a-f]{1,4})?
(?:::(?!:))
(?:(?:[0-9a-f]{1,4}(?::(?!:))){0,3}[0-9a-f]{1,4})?
|
# 0-2 quartets, wildcard, 0-5 quartets
(?:(?:[0-9a-f]{1,4}(?::(?!:))){0,1}[0-9a-f]{1,4})?
(?:::(?!:))
(?:(?:[0-9a-f]{1,4}(?::(?!:))){0,4}[0-9a-f]{1,4})?
|
# 0-1 quartets, wildcard, 0-6 quartets
(?:[0-9a-f]{1,4})?
(?:::(?!:))
(?:(?:[0-9a-f]{1,4}(?::(?!:))){0,5}[0-9a-f]{1,4})?
|
# wildcard, 0-7 quartets
(?:::(?!:))
(?:(?:[0-9a-f]{1,4}(?::(?!:))){0,6}[0-9a-f]{1,4})?
)
(?:/(?:1(?:2[0-7]|[01]\d)|\d\d?))? # With an optional CIDR routing prefix (0-128)
\s* # Trailing whitespace
$"""
我还整理了一个Python脚本来测试所有这些IPv6示例;它是here on Pastebin，因为它太大了，无法在此发布。
您可以使用“[result] = [example]”形式的测试结果和示例参数运行脚本，如下所示：
python script.py Fail=::1.2.3.4: pass=::127.0.0.1 false=::: True=::1
或者您可以通过指定无参数来简单地运行所有测试，例如：
python script.py
无论如何，我希望这有助于其他人！
答案 7 :(得分：6)
我认为这样做会......
def validIP(address):
parts = address.split(".")
if len(parts) != 4:
return False
for item in parts:
if not 0 <= int(item) <= 255:
return False
return True
答案 8 :(得分：4)
将IPv4地址视为＆＃34; ip＆＃34;。
if re.match(r'^((\d{1,2}|1\d{2}|2[0-4]\d|25[0-5])\.){3}(\d{1,2}|1\d{2}|2[0-4]\d|25[0-5])$', ip):
print "Valid IP"
else:
print "Invalid IP"
答案 9 :(得分：2)
我想出了这个noob简单版本
def ip_checkv4(ip):
parts=ip.split(".")
if len(parts)<4 or len(parts)>4:
return "invalid IP length should be 4 not greater or less than 4"
else:
while len(parts)== 4:
a=int(parts[0])
b=int(parts[1])
c=int(parts[2])
d=int(parts[3])
if a<= 0 or a == 127 :
return "invalid IP address"
elif d == 0:
return "host id should not be 0 or less than zero "
elif a>=255:
return "should not be 255 or greater than 255 or less than 0 A"
elif b>=255 or b<0:
return "should not be 255 or greater than 255 or less than 0 B"
elif c>=255 or c<0:
return "should not be 255 or greater than 255 or less than 0 C"
elif d>=255 or c<0:
return "should not be 255 or greater than 255 or less than 0 D"
else:
return "Valid IP address ", ip
p=raw_input("Enter IP address")
print ip_checkv4(p)
答案 10 :(得分：0)
我只需要解析IP v4地址。我的解决方案基于寒战策略如下：
def getIP():
valid = False
while not valid :
octets = raw_input( "Remote Machine IP Address:" ).strip().split(".")
try: valid=len( filter( lambda(item):0<=int(item)<256, octets) ) == 4
except: valid = False
return ".".join( octets )