[slurm-users] Error in job_submit.lua conditional?

[Marcus Wagner]

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.
>>>
>

-- 
Marcus Wagner, Dipl.-Inf.

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