UNITY : CLI command syntax to remove host access from lun?

Have you tried running the command without the HostID in the -lunHosts parameter ?

Hi Brett,

Yes i did try that but gives syntax error. The only way i could remove the host was by creating a fake host and assign it to the lun. This will automatically remove the existing host access. But since the GUI have the option to remove the host access i am wondering there has to be a way through CLI.

set argument with just a blank value (double or single quote) will remove all hosts.

Example:  (sv_1 is lun id):

uemcli -u admin -p pwd /stor/prov/luns/lun -id sv_1 set -lunHosts ""

uemcli -u admin -p pwd /stor/prov/luns/lun -id sv_1 set -lunHosts ","

-bash-4.2$ uemcli -d xx-unity600-01 -u admin -p mypassword /remote/host -id Host_31 set -removeLuns sv_15,sv_16,sv_17

Storage system address: xx-unity600-01

Storage system port: 443

HTTPS connection

ID = Host_31

Operation completed successfully.

-bash-4.2$

But before doing above, you grep host name that contains string "0011" to find host_ID by these 2 cli:

uemcli -d xx-unity600-01 -u admin -p mypassword /remote/host show -brief |grep  -B2 -A5 0011

27:   ID              = Host_31

      Name            = xxxx-esx0011

      Description     =

      Tenant          =

      Type            = host

      Address         =

      Netmask         =

^C

-bash-4.2$ uemcli -d xx-unity600-01 -u admin -p mypassword  /remote/host -id Host_31 show -detail

Storage system address: xx-unity600-01

Storage system port: 443

HTTPS connection

1:    ID              = Host_31

      Name            = xxxx-esx0011

      Description     =

      Tenant          =

      Type            = host

      Address         =

      Netmask         =

      OS type         = VMware ESXi 6.0.0

      Ignored address =

      Management type = Others

      Accessible LUNs = sv_15,sv_17,sv_16

      Host LUN IDs    = 0,1,2