How to "grep" out specific line ranges of a file

Question

Try using sed as mentioned on http://linuxcommando.blogspot.com/2008/03/using-sed-to-extract-lines-in-text-file.html. For example use

sed '2,4!d' somefile.txt

to print from the second line to the fourth line of somefile.txt. (And don't forget to check http://www.grymoire.com/Unix/Sed.html, sed is a wonderful tool.)

The following command will do what you asked for "extract the lines between 1234 and 5555" in someFile.

sed -n '1234,5555p' someFile

If I understand correctly, you want to find a pattern between two line numbers. The awk one-liner could be

awk '/whatev/ && NR >= 1234 && NR <= 5555' file

You don't need to run grep followed by sed.

Perl one-liner:

perl -ne 'if (/whatev/ && $. >= 1234 && $. <= 5555) {print}' file

Line numbers are OK if you can guarantee the position of what you want. Over the years, my favorite flavor of this has been something like this:

sed "/First Line of Text/,/Last Line of Text/d" filename

which deletes all lines from the first matched line to the last match, including those lines.

Use sed -n with "p" instead of "d" to print those lines instead. Way more useful for me, as I usually don't know where those lines are.

Put this in a file and make it executable:

#!/bin/bash
start=`grep -n $1 < $3 | head -n1 | cut -d: -f1; exit ${PIPESTATUS[0]}`
if [ ${PIPESTATUS[0]} -ne 0 ]; then
    echo "couldn't find start pattern!" 1>&2
    exit 1
fi
stop=`tail -n +$start < $3 | grep -n $2 | head -n1 | cut -d: -f1; exit ${PIPESTATUS[1]}`
if [ ${PIPESTATUS[0]} -ne 0 ]; then
    echo "couldn't find end pattern!" 1>&2
    exit 1
fi

stop=$(( $stop + $start - 1))

sed "$start,$stop!d" < $3

Execute the file with arguments (NOTE that the script does not handle spaces in arguments!):

1. Starting grep pattern
2. Stopping grep pattern
3. File path

To use with your example, use arguments: 1234 5555 myfile.txt

Includes lines with starting and stopping pattern.