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?
Answers:
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Method 1
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.
Method 2
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/'
Method 3
sed -e ':begin;s/pattern//2;t begin'
or without the sed goto:
sed -e 's/(pattern)/1n/;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)/1n/;h;s/.*n//3;s/pattern//g;H;g;s/n.*n//'
Method 4
A slight variation on @jillagre’s answer (modified for robustness) could look like:
sed 's/p(attern)/pn1/;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.
All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0