Something that has frustrated me in the past is using the cumbersome method of determining the connectors required to terminate a particular optic’s fibres. We would determine the optic part code (show inventory, show module, etc) and then looked up the partcode on CCO. This was incredibly inefficient and I always felt that there had to be a way of doing this through the command line so, after many CCO documents, I found the following.
show idprom int <interface>
will show the connector type.
IOS
The following is an output from a 7600. The cards in slot 1 and 2 are LAN based (67xx) cards. The ES+ in slots 3 and 4 are Ethernet Services cards, which behaves less like a classic LAN interfaces and more like a genuine router interface, with support for CBWFQ / MQC and other features commonly found on Cisco’s software router platforms, such as a 7200.
Router#sh mod Mod Ports Card Type Model Serial No. --- ----- -------------------------------------- ------------------ ----------- 1 8 CEF720 8 port 10GE with DFC WS-X6708-10GE 2 24 CEF720 24 port 1000mb SFP WS-X6724-SFP 3 2 7600 ES+ 7600-ES+2TG3CXL 4 4 7600 ES+T 76-ES+T-4TG
Some examples (filtered just to show the relevant connector lines):
Router#show idprom int tengig 1/1 | i connector Optical connector type :0x1 =SC Router#show idprom int gig 2/1 | i Connector Connector : LC [0x07] Router#show idprom int tengig 3/1 | i Connector Connector type = LC. Router#show idprom int tengig 4/1 | i Connector Connector type = LC.
CRS
CRS is (of course) different:
RP/0/RP0/CPU0:Router#show diag 0/0/CPU0 | i Connector Fri Jul 1 13:45:21.768 UTC Connector type: SC
With these commands you’ll be able to determine what your connector types are without referring to product sheets.
Noted this down for reference when this popped into my RSS Feed and stumbled across said note just now and realised I never said thank you. So thanks for the tip, very useful to know.
On a side note I noticed the “sh idprom int” output tells you very straightforwardly which ASIC a port is on which can be of use too.