bashjobslsf

How to use a variable to define bsub jobname?

I do not know the maximum number of jobs a priori. So when I keep it a variable:

#!/bin/bash
caselist=/my/caselist.txt
N=`cat $caselist | wc -l`

#BSUB -J myjob[1-$N]
...
...

(I call the above script myjob.lsf)

And submit the job as bsub < myjob.lsf, I get:

Bad job name. Job not submitted.

So is there a way I can use a variable in #BSUB -J myjob[1-$N] within myjob.lsf?

Solution

  • The code inside the file does not get evaluated when you pass it to bsub. You can probably remove it from the script entirely, and instead submit it differently.

    bsub -J "myjob[1-$(wc -l </my/caselist.txt)]" <myjob.lsf

    (speculating a bit about the bsub options; the manuals I could find online were rather bad).