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I'm trying to run:

cat s_f_d_TMT161023_1306.fits.coo.1 | sed -e '/^#/d'| sort -n -k 3 | head -n 50 >> inputcoo2.list

and,

cat s_f_d_TMT161023_1307.fits.coo.1 | sed -e '/^#/d'| sort -n -k 3 | head -n 50 >> inputcoo3.list

and so on.

The last four characters on which I'm using the cat varies from 1305 and goes till 1440. I also want the name of the output be increased by one like inputcoo3.list and inputcoo4.list and so on.

I can guess that I must run some kind of loop, but I'm new to bash and don't have any idea on how to implement it.

How do I go about it?

3

try

for i in $(seq 1305 1440)
do
  j=$((i - 1304))
  sed -e '/^#/d' "s_f_d_TMT161023_${i}.fits.coo.1" | 
    sort -n -k 3 | 
    head -n 50 >> "inputcoo${j}.list"
done
  • this save a call to cat
  • note that destination file will be appended
  • if you want leading 0 (numerical zero, not character o) in filename, use j="$(printf '%03d' $i)"
  • 'seq' has a bash builtin of {$start..$end} - You could replace the first line with 'for i in {1305..1440}' – xrobau Jul 25 '17 at 14:09
  • 1
    @xrobau thanks, I didn't know it. seq didn't exists when I start using unix. – Archemar Jul 25 '17 at 14:11
  • No worries. Bash has so many useful builtins that people don't know exist 8) – xrobau Jul 25 '17 at 14:23

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