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I have a logfile with timestamps in it. Occasionally there are multiple timestamps in one line. Now I would like to remove all of the timestamps from a line but keep the first one.
I can do s/pattern//2 but that only removes the second occurrence and sed doesn't allow something like s/pattern//2-.
Any suggestions?
With GNU sed:
sed 's/pattern//2g'
The 2 specifies that the second pattern and all the restg should remove. So this will keep the first one.
sed: 1: "s/^ *$//2g": more than one number or 'g' in substitute flags - with gsed, it would be as if nothing happened.
This should work (replace _ by something else should it clash with your logs):
sed -e 's/pattern/_&/1' -e 's/\([^_]\)pattern//g' -e 's/_\(pattern\)/\1/'
sed -e ':begin;s/pattern//2;t begin'
or without the sed goto:
sed -e 's/\(pattern\)/\1\n/;h;s/.*\n//;s/pattern//g;H;g;s/\n.*\n//'
The generic solutions to remove from the nth (3 for example) position are:
sed -e ':begin;s/pattern//4;t begin'
sed -e 's/\(pattern\)/\1\n/;h;s/.*\n//3;s/pattern//g;H;g;s/\n.*\n//'
A slight variation on @jillagre's answer (modified for robustness) could look like:
sed 's/p\(attern\)/p\n\1/;s///g;s/\n//'
...but in some seds you may need to replace the n in the right-hand side of the first s///ubstitution statement with a literal \newline character.
