我尝试用pysnmp启动SNMP代理。对于IPv4和IPv6绑定,它与localhost("27.0.0.1"one_answers"::1")配合使用非常好
但当我尝试使用从接口获取的其他IPv6 IP时,由于,它失败了
[vagrant@test SOURCES]$ sudo python snmp_agent.py enp0s8
Traceback (most recent call last):
File "snmp_agent.py", line 172, in <module>
master_agent_startup(ifname=sys.argv[1])
File "snmp_agent.py", line 101, in master_agent_startup
(get_ipv6_address(interface_name), SNMP_AGENT_PORT))
File "/usr/lib/python2.7/site-packages/pysnmp/carrier/asyncore/dgram/base.py", line 50, in openServerMode
raise error.CarrierError('bind() for %s failed: %s' % (iface, sys.exc_info()[1],))
pysnmp.carrier.error.CarrierError: bind() for ('fe80::a00:27ff:fe9e:9c16', 8001) failed: [Errno 22] Invalid argument
这是接口"enp0s8"的输出:
[vagrant@test SOURCES]$ ifconfig enp0s8
enp0s8: flags=4163<UP,BROADCAST,RUNNING,MULTICAST> mtu 1500
inet 172.20.20.26 netmask 255.255.255.0 broadcast 172.20.20.255
inet6 fe80::a00:27ff:fe9e:9c16 prefixlen 64 scopeid 0x20<link>
ether 08:00:27:9e:9c:16 txqueuelen 1000 (Ethernet)
RX packets 874053 bytes 115842841 (110.4 MiB)
RX errors 0 dropped 0 overruns 0 frame 0
TX packets 862314 bytes 114652475 (109.3 MiB)
TX errors 0 dropped 0 overruns 0 carrier 0 collisions 0
这是我用于IPv6绑定的代码段:
def get_ipv6_address(ifname):
return netifaces.ifaddresses(ifname)[netifaces.AF_INET6][0]['addr'].split('%')[0]
config.addSocketTransport(snmpEngine, udp.domainName,
udp.UdpTransport().openServerMode(
(get_ipv4_address(interface_name), SNMP_AGENT_PORT))
)
config.addSocketTransport(snmpEngine, udp6.domainName,
udp6.Udp6SocketTransport().openServerMode(
(get_ipv6_address(interface_name), SNMP_AGENT_PORT))
)
从pysnmp示例来看,"openServerMode()"中的参数似乎只是IP和端口的元组。根据输出错误,我想给定的IP和端口并没有错误。那么,为什么它由于无效参数而失败呢?你@Ilya Etingof或其他pysnmp专家能帮我吗?
谢谢。
更新:我试图将它与给定的建议绑定,但仍然不起作用。bind命令是在一个新安装的CentOS上运行的。但它仍然失败了:[root@test~]#ifconfig
eth1: flags=4163<UP,BROADCAST,RUNNING,MULTICAST> mtu 1500
inet 10.10.20.4 netmask 255.255.255.0 broadcast 10.10.20.255
inet6 fe80::f816:3eff:fee1:5475 prefixlen 64 scopeid 0x20<link>
ether fa:16:3e:e1:54:75 txqueuelen 1000 (Ethernet)
RX packets 12242 bytes 962552 (939.9 KiB)
RX errors 0 dropped 0 overruns 0 frame 0
TX packets 12196 bytes 957826 (935.3 KiB)
TX errors 0 dropped 0 overruns 0 carrier 0 collisions 0
[root@test ~]# python
Python 2.7.5 (default, Oct 11 2015, 17:47:16)
[GCC 4.8.3 20140911 (Red Hat 4.8.3-9)] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import socket
>>> s = socket.socket(socket.AF_INET6, socket.SOCK_DGRAM, 0)
>>> addr_and_port = ('fe80::f816:3eff:fedb:ba4f', 8001)
>>> s.bind(addr_and_port)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/usr/lib64/python2.7/socket.py", line 224, in meth
return getattr(self._sock,name)(*args)
socket.error: [Errno 22] Invalid argument
>>>
[1]+ Stopped python
[root@test ~]# netstat -anp | grep 8001
[root@test ~]#
还有一个更新:我想绑定失败是因为我的环境在IPv6配置方面有一些问题。因为我只能通过使用"socket.getaddrinfo()"方法获得一个IPv4地址。
Br,-大鹏角
如果多次尝试绑定同一套接字,可能会出现此错误。但我看不出你的代码是这样的。
.openServerMode()
方法并没有什么神奇之处——它只是在套接字对象上调用.bind()
。为了获得灵感,这在Python提示下是否有效?
from pysnmp.carrier.asyncore.dgram import udp6
addr_and_port = ('fe80::a00:27ff:fe9e:9c16', 8001)
udp6.Udp6SocketTransport().openServerMode(addr_and_port)
甚至:
import socket
s = socket.socket(socket.AF_INET6, socket.SOCK_DGRAM, 0)
addr_and_port = ('fe80::a00:27ff:fe9e:9c16', 8001)
s.bind(addr_and_port)
我希望这样的测试能帮助你解决问题。。。
当您使用IPv6链接本地地址时,您必须始终将作用域与它一起使用。如果没有作用域,链接本地地址将无效,因此您将收到Invalid argument
错误。
例如,您必须使用fe80::a00:27ff:fe9e:9c16%enp0s8
,而不是使用fe80::a00:27ff:fe9e:9c16
。