Free BSD/ Mac OS X Sed: Print regex match and the line 5 lines after the match

sedtext processing

I thought it would be sed -n /mymatch/,+5p but I get expected context address. At this point I am just guessing commands, nothing works! I want to grab every line that matches the regex and print that line and the line which comes 5 lines after it, globally from a file.

Best Answer

Might be easier with awk:

awk '/foo/ {print; p[NR+5]; next}; NR in p'

Related Solutions

Extracting subset of lines of a file based on regex for first & last line

With sed:

sed -n '/R/,$!d; /S/q; p'

Example:

$ seq 20 | sed -n '/6/,$!d; /1/q; p'
6
7
8
9

Print match and line after

How about:

awk '$4=="-" && prev4=="frz" {print prevline; print} {prev4 = $4; prevline=$0}' file

Best Answer

Related Solutions

Extracting subset of lines of a file based on regex for first & last line

Print match and line after

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