Shell – Best way to cancel all the SLURM jobs from shell command output

shellslurm

I submitted lots of SLURM job script with debug time limit (I forgot to change the time for actual run). Now they are all submitted at the same time, so they all start with job ID 197xxxxx. Now, I can do

squeue -u $USER | grep 197 | awk '{print $1}'

to print the job ID's I want to delete. But how do I use scancel command on all these ID's. The output from the above shell command would look like

Best Answer

squeue -u $USER | grep 197 | awk '{print $1}' | xargs -n 1 scancel

Check the documentation for xargs for details. If scancel accepts multiple job ids (it should), you may omit the -n 1 part.

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You seem to want the count instead of the 'Start' output.

Display count instead:

tcpdump -i any -s 65535 host 192.168.1.110 and port 1645 or port 1813 -v -X |
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Display count and line:

tcpdump -i any -s 65535 host 192.168.1.110 and port 1645 or port 1813 -v -X |
  awk '/Start/ {c++; print c, $0} {}'

Display count on the same line:

tcpdump -i any -s 65535 host 192.168.1.110 and port 1645 or port 1813 -v -X |
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Shell – Print first and last line of all files in folder

So I like sed the answer can be

for file in file.log.*
do
   echo "file: $file"
   echo -n "first line: "
   cat "$file" | sed -n '/^\s*$/!{p;q}'
   echo -n "last line: "
   tac "$file" | sed -n '/^\s*$/!{p;q}'
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Best Answer

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