[file] Printing the last column of a line in a file

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?

One way using awk:

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

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

awk '{print $(NF)}' 

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

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

$

You can do this without awk with just some pipes.

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

Execute this on the file:

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

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

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.

You can do all of it in awk:

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

maybe this works?

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

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.