lmutil lmremove - syntax and execution 1

[Lockdain]

Hello All!

I'm trying to remove a single user's license for a specified feature. Flexlm guide offers such a syntax for lmremove:

lmremove [ -c file ] feature user host display
-c license_file
Use the license file named. If this switch is not specified, lmgrd looks for the environment variable LM_LICENSE_FILE. If the environment variable is not set, lmgrd looks for the default license file.
feature
The name of the feature checked out by the user.
user
The name of the user whose license you are removing.
host
The name of the host the user is logged in to.
display
The name of the display where the user is working.
​

The following information i got from the lmutil lmstat -a output:

feature;
user;
host;
license_file;
​

The main question is about what does the display option means. I can't understand it clearly.
That what the FAQ says:

display
The name of the display where the user is working.​

But i can't find such parameter in lmstat output. Which parameter i have to define?

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[apekas]

From the lmutil lmstat -a command, you should get an output like this

[pre]joe nirvana nirvana (v1.000) (cloud9/7654 102), start Fri 10/29 18:40[/pre]

from this output:

user = joe
user_host = nirvana
display = nirvana
server_host = cloud9
TCP/IP port = 7654
license handle = 102

the lmremove syntax can be either of these two:

lmremove [-c license_file] feature user user_host display

lmremove [-c license_file] -h feature server_host port handle

 

[Lockdain]

Hi @apekas!

Appreciate you for the help. The problem was in my Windows command prompt - it failed to show me the whole lmstat -a report and i wasn't able to get all needed information.

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