Bash: run script in background and write output to log

Question

I want to run script in background and write it output to log:

Here is my original script:

for i in `seq 1 3`; do ./script.sh $i &> "my_logs/"$i".log"; done

What I have tried:

for i in `seq 1 3`; do ./script.sh $i &> "my_logs/"$i".log" &; done

What is a proper way to do this?

for i in `seq 1 3`; do ./script.sh $i & &> "my_logs/"$i".log"; done

What is wrong with your attempt? (Aside from the unnecessary leaving-out of $i from the quotes: "my_logs/$i.log" is fine.) — chepner
– chepner, Commented Mar 1, 2017 at 15:10

arco444 · Accepted Answer · 2017-03-01 15:11:54Z

1

I think you want:

for i in `seq 1 3`; do ./script.sh $i > "my_logs/$i.log"; done &

answered Mar 1, 2017 at 15:11

arco444

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1 Comment

I need each ./script.sh to be runned as separate background process and this seems just run for loop in background?