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 fi

Answer 1

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

$

Answer 2

maybe this works?

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

Answer 3

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.

Answer 4

You can do all of it in awk:

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

Answer 5

One way using awk:

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

Answer 6

You can do this without awk with just some pipes.

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