Using awk, I would like to print the last matching line of a file. I would like only the matching line itself, not any range of lines. I can use a command like this
awk '/foo/' bar.txt However this will print all matching lines, not just the last one.
You can save the value in a variable and then print it after processing the whole file:
awk '/foo/ {a=$0} END{print a}' file $ cat file foo1 foo2 foo3 4 5 $ awk '/foo/ {a=$0} END{print a}' file foo3 
reverse the file and find the first match.
tac file | awk '/foo/ {print; exit}' 
tac file | grep -m1 foo so that MAY be slightly better as it's briefer and IMHO no less clear. Won't work with all greps though, it might be GNU grep specific.
tac $file | grep $pattern | tac | tail -n $N -- to print last N matched lines preserving ordersed solution:
sed -n '/foo/h;$!b;g;p' bar.txt Place the matching regex in hold buffer and branch out. Keep doing this until end of file. When the end of file is reached, grab the line from hold space and place it on pattern space. Print pattern space.
The command tail allows you to retreive the last n lines of an input stream. You can pipe your awk output into tail in order to trim off everything except the last line:
awk '/foo/' bar.txt | tail -1 
END.You can assign values to variables and print them at the end of processing both the lines and their number.
File:
$ cat file foo1 foo2 foo3 4 5 Command:
$ awk '/foo/ {a=$0;n=NR} END{print n, a}' file 3 foo3 