Question

如何通过SNMP获得此结果？

显示etherchannel摘要

Group  Port-channel  Protocol    Ports
------+-------------+-----------+-----------------------------------------------
1      Po1(RU)          -        Gi2/3(P)       Gi3/3(P)       
2      Po2(RU)          -        Gi2/4(P)       Gi3/4(P)       
3      Po3(SU)          -        Gi2/2(P)       Gi3/2(P)       
9      Po9(SU)          -        Gi2/12(P)      Gi3/12(P)      
421    Po421(SU)        -        Gi11/1(P)      Gi11/2(P)      Gi11/3(P)      
                             Gi11/4(P)      
422    Po422(SU)        -        Gi11/5(P)      Gi11/6(P)      Gi11/7(P)      
                             Gi11/8(P)      
423    Po423(SU)        -        Te12/1(P)

Answer 1

这有点晚了，但研究类似的事情我遇到this Cisco forums post。它在IEEE8023-LAG-MIB中报告以下对象：

IEEE8023-LAG-MIB::dot3adAggPortSelectedAggID.28 = INTEGER: 337
IEEE8023-LAG-MIB::dot3adAggPortSelectedAggID.82 = INTEGER: 337

IEEE8023-LAG-MIB::dot3adAggPortAttachedAggID.28 = INTEGER: 337
IEEE8023-LAG-MIB::dot3adAggPortAttachedAggID.82 = INTEGER: 337

ifName.28 = Gi2/4
ifName.82 = Gi3/10
ifName.337 = Po1

在同一环境中，show etherchannel summary命令将导致：

+------+-------------+-----------+---------------------------------+
| Group|Port-channel |Protocol   | Ports                           |
+------+-------------+-----------+---------------------------------+
|  1   | Po1(SU)     | LACP      | Gi2/4(P)       Gi3/10(P)        |
+------+-------------+-----------+---------------------------------+

如何通过SNMP将端口通道映射到接口

1 个答案: