Printing the last column of a line in a file

Question

I have a file that is constantly being written to/updated. I want to find the last line containing a particular word, then print the last column of that line.

The file looks something like this. More A1/B1/C1 lines will be appended to it over time.

A1 123 456
B1 234 567
C1 345 678
A1 098 766
B1 987 6545
C1 876 5434

I tried to use

tail -f file | grep A1 | awk '{print $NF}'

to print the value 766, but nothing is output.

Is there a way to do this?

Pervious

Answer 1

You don't see anything, because of buffering. The output is shown, when there are enough lines or end of file is reached. tail -f means wait for more input, but there are no more lines in file and so the pipe to grep is never closed.

If you omit -f from tail the output is shown immediately:

tail file | grep A1 | awk '{print $NF}'

---

@EdMorton is right of course. Awk can search for A1 as well, which shortens the command line to

tail file | awk '/A1/ {print $NF}'

or without tail, showing the last column of all lines containing A1

awk '/A1/ {print $NF}' file

---

Thanks to @MitchellTracy's comment, tail might miss the record containing A1 and thus you get no output at all. This may be solved by switching tail and awk, searching first through the file and only then show the last line:

awk '/A1/ {print $NF}' file | tail -n1

Harmonica

You never need grep+awk since awk can do it's own RE comparison. "grep A1 | awk '{print $NF}'" should just be "awk '/A1/{print $NF}'". Also, if you aren't using tail -f then you don't need tail at all, yoiu can just do: awk '/A1/{f=$NF} END{print f}' file – Granitite

This isn't quite right, as the column may not appear in the window that raw tail gives you. Unless you know it is going to appear with a certain frequency, it would be safer to awk '/A1/ {print $NF}' file | tail -n1. – Westernize

Answer 2

To print the last column of a line just use $(NF):

awk '{print $(NF)}' 

Matriculation

Why not just $NF? – Catcher

Answer 3

One way using awk:

tail -f file.txt | awk '/A1/ { print $NF }'

Suavity

I thought that was the right solution to the posted problem (+1!) but now I see I don't understand the question. The OP wants the last field from the LAST LINE of the file, but there is no LAST LINE if you're using tail -f. Maybe he means the last line CURRENTLY in the file, in which case see one of my other comments. – Granitite

Very good and simple. Thumbs up. Thumbs down for awk's cryptic syntax :) – Ops

awk's syntax might appear cryptic nowadays, but it was really ahead of its time. In 1977 (when awk was born), C was a kindergartener (at 5yo), and C++ a glimmer in Bjarne's eye. It's the only language I use where it's normal to include your entire program as a command-line parameter! I might be biased, though, since I've always been a bit awk-ward. ;) – Kinnikinnick

Answer 4

You can do this without awk with just some pipes.

tac file | grep -m1 A1 | rev | cut -d' ' -f1 | rev

Solutrean

This solves your problem, but it doesn't really use awk. Hope it's good enough for you anyway. – Solutrean

Answer 5

maybe this works?

grep A1 file | tail -1 | awk '{print $NF}'

Furfur

Answer 6

Not the actual issue here, but might help some one: I was doing awk "{print $NF}", note the wrong quotes. Should be awk '{print $NF}', so that the shell doesn't expand $NF.

Galleass

Answer 7

You can do all of it in awk:

<file awk '$1 ~ /A1/ {m=$NF} END {print m}'

Sesame

You don't need split() and an array, just save the last field and print that: awk '/A1/{f=$NF} END{print f}' file – Granitite

Answer 8

Using Perl

$ cat rayne.txt
A1 123 456
B1 234 567
C1 345 678
A1 098 766
B1 987 6545
C1 876 5434


$ perl -lane ' /A1/ and $x=$F[2] ; END { print "$x" } ' rayne.txt
766

$

Scriven

Looks too complicated, just do perl -lane 'print $F[2] if /A1/' rayne.txt – Serles

@Hi-Angel..the question asks for last occurrence of A1. your answer will print all matches.. btw if you like my answer, pls upvote it – Scriven

Answer 9

awk -F " " '($1=="A1") {print $NF}' FILE | tail -n 1

Use awk with field separator -F set to a space " ".

Use the pattern $1=="A1" and action {print $NF}, this will print the last field in every record where the first field is "A1". Pipe the result into tail and use the -n 1 option to only show the last line.

Cockcrow

Answer 10

Execute this on the file:

awk 'ORS=NR%3?" ":"\n"' filename

and you'll get what you're looking for.

Samoyed

Answer 11

ls -l | awk '{print $9}' | tail -n1

Gamelan

How is this even related to the OP's question? – Whaleback