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?

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

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

awk '{print $(NF)}' 

One way using awk:

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

You can do this without awk with just some pipes.

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