[SOLVED] How to get lines from the last match to the end of file?

Issue

Need to print lines after the last match to the end of file. The number of matches could be anything and not definite. I have some text as shown below.

MARKER
aaa
bbb
ccc
MARKER
ddd
eee
fff
MARKER
ggg
hhh
iii
MARKER
jjj
kkk
lll

Output desired is

jjj
kkk
lll

Do I use awk with RS and FS to get the desired output?

Solution

You can actually do it with awk (gawk) without using any pipe.

$ awk -v RS='(^|\n)MARKER\n' 'END{printf "%s", $0}' file
jjj
kkk
lll

Explanations:

• You define your record separator as (^|\n)MARKER\n via RS='(^|\n)MARKER\n', by default it is the EOL char
• 'END{printf "%s", $0}' => at the end of the file, you print the whole line, as RS is set at (^|\n)MARKER\n, $0 will include all the lines until EOF.

---

Another option is to use grep (GNU):

$ grep -zoP '(?<=MARKER\n)(?:(?!MARKER)[^\0])+\Z' file
jjj
kkk
lll

Explanations:

• -z to use the ASCII NUL character as delimiter
• -o to print only the matching
• -P to activate the perl mode
• PCRE regex: (?<=MARKER\n)(?:(?!MARKER)[^\0])+\Z explained here https://regex101.com/r/RpQBUV/2/

---

Last but not least, the following sed approach can also been used:

sed -n '/^MARKER$/{n;h;b};H;${x;p}' file
jjj
kkk
lll

Explanations:

• n jump to next line
• h replace the hold space with the current line
• H do the same but instead of replacing, append
• ${x;p} at the end of the file exchange (x) hold space and pattern space and print (p)

that can be turned into:

tac file |  sed -n '/^MARKER$/q;p' | tac

if we use tac.

Answered By - Allan

Answer Checked By - David Marino (WPSolving Volunteer)