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I use the command

varioutput=$(awk '{print $j}' OFS=, data/damper.test_temp1.csv)

because I want to extract the j th value of a line of many values seperated by , . But when I want to use $varioutput it gives me always the whole line.

What am I doing wrong?

Actually I want to use

for ((j=1; j<=20; j++)); do
         varioutput=$(awk -F, -v jj="$j" '{print $jj}'  data/damper.test_temp1.csv)
done

So I am quite confused now weather I should use it like above or not? I get an error, but when I use it like

for ((j=1; j<=20; j++)); do
         varioutput=$(awk -F, '{print $j}'  data/damper.test_temp1.csv)
done

I still get the line.

Now the following was the solution

for ((j=1; j<=20; j++)); do
         varioutput=$(awk -F, -v k=$j '{print $k}'  data/damper.test_temp1.csv)
done
  • What error do you get? – Jesse_b May 9 at 14:27
2

You’re specifying the output field separator, not the input field separator; use this instead:

varioutput=$(awk -F, '{print $j}' data/damper.test_temp1.csv)

(or set FS instead of OFS).

I’m also assuming that j is a placeholder above, and that you’re replacing it statically with the appropriate value (for example, print $4).


To use another variable from script, you must pass it to awk

e.g. for RANK variable in shell

varioutput=$(awk -F, -v j=$RANK '{print $j}' data/damper.test_temp1.csv)

Generally speaking, if you start using AWK for small pieces of a script as in this example, it’s better to use AWK for more of the script.

  • @steeldriver I think j is a placeholder here, the real command would be print $3 or something like that. – Stephen Kitt May 9 at 13:56
  • ah yes that makes sense – steeldriver May 9 at 13:57
  • I add way to do it in case $j should be a real variable. – Archemar May 9 at 14:29

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