How to grep the last occurrence of a line pattern

Question

I have a file with contents

x
a
x
b
x
c

I want to grep the last occurrence,

x
c

when I try

sed -n  "/x/,/b/p" file

it lists all the lines, beginning x to c.

Solution 1

I'm not sure if I got your question right, so here are some shots in the dark:

• Print last occurence of x (regex):

  grep x file | tail -1
  

• Alternatively:

  tac file | grep -m1 x
  

• Print file from first matching line to end:

  awk '/x/{flag = 1}; flag' file
  

• Print file from last matching line to end (prints all lines in case of no match):

  tac file | awk '!flag; /x/{flag = 1};' | tac
  

Solution 2

grep -A 1 x file | tail -n 2

-A 1 tells grep to print one line after a match line
with tail you get the last two lines.

or in a reversed way:

tac fail | grep -B 1 x -m1 | tac

Note: You should make sure your pattern is "strong" enough so it gets you the right lines. i.e. by enclosing it with ^ at the start and $ at the end.

Solution 3

This might work for you (GNU sed):

sed 'H;/x/h;$!d;x' file

Saves the last x and what follows in the hold space and prints it out at end-of-file.

Solution 4

not sure how to do it using sed, but you can try awk

awk '{a=a"\n"$0; if ($0 == "x"){ a=$0}}  END{print a}' file

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