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

问题

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?

回答1:

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.

回答2:

Could you please try following.

tac file | awk '/MARKER/{print val;exit} {val=(val?val ORS:"")$0}' | tac

Benefit of this approach will be awk will just read last block of the Input_file(which will be actually first block for awk after tac prints it reverse)and exit after that.

Explanation:

tac file |                      ##Printing Input_file in reverse order.
awk '
  /MARKER/{                     ##Searching for a string MARKER in a line of Input_file.
    print val                   ##Printing variable val here. Because we need last occurrence of string MARKER,which has become first instance after reversing the Input_file.
    exit                        ##Using exit to exit from awk program itself.
  }
  {
    val=(val?val ORS:"")$0      ##Creating variable named val whose value will be keep appending to its own value with a new line to get values before string MARKER as per OP question.
  }
' |                             ##Sending output of awk command to tac again to make it in its actual form, since tac prints it in reverse order. 
tac                             ##Using tac to make it in correct order(lines were reversed because of previous tac).

回答3:

You can try Perl as well

$ perl -0777 -ne ' /.*MARKER(.*)/s and print $1 ' input.txt

jjj
kkk
lll

$

回答4:

This might work for you (GNU sed):

sed -nz 's/.*MARKER.//p' file

This uses greed to delete all lines upto and including the last occurrence of MARKER.

回答5:

Simplest to remember:

tac fun.log | sed "/MARKER/Q" | tac

来源：https://stackoverflow.com/questions/56334150/how-to-get-lines-from-the-last-match-to-the-end-of-file