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?

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

Answer 2

要打印一行的最后一列，只需使用 $(NF)：

awk '{print $(NF)}' 

Answer 3

使用awk一种方法：

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

Answer 4

你可以在没有 awk 的情况下使用一些管道来做到这一点。

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

Answer 5

也许这有效？

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

Answer 6

您可以在awk完成所有操作：

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

Answer 7

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 8

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 9

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 .

Answer 10

Execute this on the file:

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

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

Answer 11

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