Print specific range of lines between two matches starting with a specific line after the first match

Question

How can I print all data between two matches, staring from specific line after the first match. For example, let's say I have two patterns Linux and Unix and there are entries between these two words I would like to print data from specific line after the word Linux, in the example below line #6 after Linux, until the word Unix:

Linux
12
32
34
556
676
77
85434
rdb
232323
232435656
Unix
98736
4345
4546456
456

expected output:

85434
rdb
232323
232435656

How can I do it using awk, sed and grep?

Answer 1

Whenever you see relative addressing, you should reach for ed.

echo '/Linux/+7,/Unix/-1p' | ed -s input_file.txt -

Or, as a heredoc:

ed -s input_file.txt - <<\EOF
/Linux/+7,/Unix/-1p
EOF

Answer 2

Try this,

awk '/^Linux$/{flag=1;next}/^Unix$/{flag=0}flag' file | tail -n +7

Or without tail:

awk '/^Linux$/{flag=1;n=NR;next}/^Unix$/{flag=0};flag&&NR>(n+6);' file

Explanation:

• Search for pattern ^Linux$, and if found, set flag to 1 (true) and move directly to next line.
• Go on until pattern ^Unix$ was found and set flag to 0 (false)
• The flag at the end is an awk specific notation. If flag is true, print the line, else do nothing.
• flag is just a variable name and can be anything, e.g. f or printthisline.

---

Using Gnu sed:

sed '/^Linux$/,/^Unix$/{//!b};d' file | tail -n +7

Answer 3

I would write one of

sed -n '/Linux/,/Unix/p' file | sed '1,7d;$d'
# or
awk '/Linux/ {nr = NR} /Unix/ {exit} nr && NR > nr+6' file

Answer 4

if you search with grep you can do this:

• grep -A 5 "string used to search" <path-to-file(s)> will show the 5 lines after the matched string
• grep -B 5 "string used to search" <path-to-file(s)> will show the 5 lines before the matched string
• grep -C 5 "string used to search" <path-to-file(s)> will show the 5 lines after and before the matched string (C stands for context)

for your example, it would be: grep -A 2 -B 1 "rdb" <path-to-file(s)>