Print non consecutive lines from file given line number list

Question

Consider that after grep, awk and grep again I have a list of line numbers looking like this:

20
55
98
154
1100
...

The command I'm using looks like this:

grep '01/01' /var/log/some.log | grep 'some' | awk '{print $10}' | grep -vn 200 | awk -F ':' '{print $1}'

I would like to now print those lines from that file, ideally by continue to pipe this command. Any ideas?

I know I could simply remove awk '{print $10}'and awk -F ':' '{print $1}' but for some reason when I do this I get no results, whereas when I try with the command posted above I do get the list of numbers.

Answer 1

Solution:

| xargs -I{} sed '{}q;d' /var/log/some.log

See How to get n-th line from a file

Comment:

You could probably shorten your long sequence of pipes (including the above last step) by writing a more elaborate sed, awk or perl command:

perl -lane 'print $F[9] if /01\/01/ and /some/ and $F[9]!~/200/' /var/log/some.log