Thanks!
Prentice
On 2/6/19 11:00 AM, Marcus Wagner wrote:
Hi Prentice,
there, I might help.
I've created a table, e.g.:
local userflags = {
-- "<username>" = {
-- "bypass" = 1, # optional, if you want to bypass the
submit_plugin
-- "debug" = 1, # optional, if you want to get debug
messages
-- "param" = 1, # optional, if you want the parameters be
printed
-- "submitall" = 1, # optional, if you should be allowed to
submit to all projects
-- "nosubmit" = 1, # optional, if you should be prohibited
to submit at all
-- "ignorequota" = 1, # optional, if the quota should be ignored
-- }
mw445520 = { bypass = 0, debug = 0, param = 1, submitall = 0,
nosubmit = 0, ignorequota = 0, },
}
and a debug function, which checks if for username the dub entry is
set to 1. It then prints the debug message.
So you can use something like
debug("This is a test message")
and only the users, whose debug flag is set, see this message.
As long as you use "debug" for debugging messages, the "normal" users
won't get disturbed.
Best
Marcus
On 2/6/19 4:04 PM, Prentice Bisbal wrote:
"Dirty debugging" I like that. I'm going to use that from now on. I
have tried that method in the past while debugging other issues. I
try not to use it too much, since I don't want these "dirty
debugging" messages being seen by users (I don't have a test
environment, so I have to test debug in production), but it looks
like I may have to do that.
Prentice
On 2/5/19 9:00 AM, mercan wrote:
Hi;
I think dirty debugging is required using printf (slurm.log_user),
because the lua of our slurm installation returns a lot of variables
as nil. You can limit the output to a specific user as below:
if job_desc.user_name == "mercan" then
slurm.log_user("job_desc.user_id=")
slurm.log_user(job_desc.user_id)
slurm.log_user("job_desc.partition=")
slurm.log_user(job_desc.partition)
end
Ahmet M.
On 5.02.2019 01:27, Prentice Bisbal wrote:
Can anyone see an error in this conditional in my job_submit.lua?
if ( job_desc.user_id == 28922 or job_desc.user_id == 41266 )
and ( job_desc.partition == 'general' or job_desc.partition ==
'interruptible' ) then
job_desc.qos = job_desc.partition
return slurm.SUCCESS
end
I am one if those user id's but if I submit a job to partition
'interruptible', without specifying a QOS, it still gets assigned
to the default QOS, which is 'general':
cat mpihello.sbatch
#!/bin/bash
#SBATCH -n 32
#SBATCH -p interruptible
#SBATCH -t 00:01:00
#SBATCH -J mpihello
#SBATCH -o mpihello-%j.out
#SBATCH -e mpihello-%j.err
#SBATCH --mail-type=ALL
module load gcc/7.3.0
module load openmpi/3.0.0
srun --mpi=pmi2 ./mpihello
$ scontrol show job 433953 | grep QOS
Priority=2512 Nice=0 Account=unix QOS=general
The logic of that conditional seems pretty simple, and I'm using
similar compound conditionals throughout my job_submit.lua script.
Can't figure out where the mistake is in this one.
