sed replace last line matching pattern

Question

Given a file like this:

a
b
a
b

I\'d like to be able to use sed to replace just the last line that contains an instance of \"a

Answer 1

awk-only solution:

awk '/a/{printf "%s", all; all=$0"\n"; next}{all=all $0"\n"} END {sub(/^[^\n]*/,"c",all); printf "%s", all}' file

Explanation:

• When a line matches a, all lines between the previous a up to (not including) current a (i.e. the content stored in the variable all) is printed
• When a line doesn't match a, it gets appended to the variable all.
• The last line matching a would not be able to get its all content printed, so you manually print it out in the END block. Before that though, you can substitute the line matching a with whatever you desire.

Answer 2

Here is a way with only using awk:

awk '{a[NR]=$1}END{x=NR;cnt=1;while(x>0){a[x]=((a[x]=="a"&&--cnt==0)?"c <===":a[x]);x--};for(i=1;i<=NR;i++)print a[i]}' file

$ cat f
a
b
a
b
f
s
f
e
a
v
$ awk '{a[NR]=$1}END{x=NR;cnt=1;while(x>0){a[x]=((a[x]=="a"&&--cnt==0)?"c <===":a[x]);x--};for(i=1;i<=NR;i++)print a[i]}' f
a
b
a
b
f
s
f
e
c <===
v