Vtyjkyuk

I have a file which contains multiple occurrences of pattern "====". These patterns are followed by texts. I'd like to search for the last occurrence of the "====" pattern and print all the lines after the pattern. Any ideas on how to do it in bash? Combination of grep, awk, sed or tail, etc..?

Sample file:

====

First run of script

End of first Script

====

2nd run of script

End of 2nd script

====

3rd run of script

End of 3rd script

Output of the command should look like below.

3rd run of script

End of 3rd script

Based on SO duplicate:

$ sed -n '/====/h;//!H;$!d;x;//p' sample.txt 
====

3rd run of script

End of 3rd script

Output of the command should look like below.

3rd run of script

End of 3rd script

I started writing an explanation, but I don't actually understand some of the commands so I'll just refer to info sed.

Software tools method, more efficient for big files since it only reads the end of the file, then stops when it finds the last match:

tac sample.txt | grep -F -m1 -B 999999 '====' | head -n -1 | tac

Note: increase or reduce 999999 as needed, just so it's longer than any possible match. See also *Glenn Jackman's answer with variants for awk and sed which avoid the need for 99999. For systems with low resources, the grep method is the most efficient of the three variants.

save the text in file.txt and run this command:

awk 'p; /====/ $p' file.txt | tail -3

output:

3rd run of script

End of 3rd script

Same concept as agc's answer, but doesn't require you to guess how many lines there might be:

with GNU sed

tac file | sed '/====/Q' | tac

or awk

tac file | awk '/====/ exit 1' | tac

The idea of the double tac is to invert the question: How to print all lines preceding the first occurrence of pattern. That is a much simpler problem.

In awk, something like this:

awk '/====/ t = ""; next t = t $0 "n"; END printf "%s", t; ' < sample.txt

It stores the input lines in t, and clears t when it sees ====. What ever is in there is printed at the end.

Note that

• The ==== doesn't have to be on a line of its own (strictly speaking, you didn't say it should be). If you want to only recognize it when it's the only thing on a line, use /^====$/ instead.


• This prints the whole input if ==== is not seen. Checking that there is at least one needs an extra condition, see below.


• Your sample output is missing the first empty line after the ====, but that might be just the post formatting. If you do want to remove it, add a pipe to sed 1d.

Another version that only starts storing input after the ==== separator is seen, effectively producing an empty output if the input doesn't contain the separator:

awk '/====/ any = 1; t = ""; next any t = t $0 "n"; END printf "%s", t; ' < sample.txt