Checking if a Screen of the Specified Name Exists

Question

I have made a bash file which launches another bash file in a detached screen with a unique name, I need to ensure that only one instance of that internal bash file is running at any one point in time. To do this, I want to make the parent bash file check to see if the screen by that name exists before attempting to create it. Is there a method to do this?

Answer 1

You can grep the output of screen -list for the name of the session you are checking for:

if ! screen -list | grep -q "myscreen"; then
    # run bash script
fi

Answer 2

You can query the screen 'select' command for a particular session; the shell result is '0' if the session exists, and '1' if the named screen session is not found:

$ screen -S Tomcat
$ screen -S Tomcat -Q select . ; echo $?
0

versus:

$ screen -S Jetty -Q select . ; echo $?
No screen session found.
1

Note that the '.' after the select is optional, but may be more robust.

Answer 3

Given I can't comment I am posting this as a new answer. troyfolger's answer is a good idea and basically amounts to try and send the session a command that will do very little. One issue is that for some (older) versions of screen -Q isn't supported and so for those versions the correct command is

screen -S Jetty -X select . ; echo $?

Which send the command "select ." to the screen session called "Jetty".

Select changes which window is active and . means the current active window so this means try and change the active window to the currently active window. This can only fail if there isn't a session to connect to which is what we wanted.

If you read the info docs than it does suggest that the only use of select . is with -X as a test or to make sure something is selected.