How do I get the last non-empty line of a file using tail in Bash?

Question

How do I get the last non-empty line using tail under Bash shell?

For example, my_file.txt looks like this:

hello
hola
bonjour
(empty line)
(empty line)

Obviously, if I do tail -n 1 my_file.txt I will get an empty line. In my case I want to get bonjour. How do I do that?

Answer 1

You can use Awk:

awk '/./{line=$0} END{print line}' my_file.txt

This solution has the advantage of using just one tool.

Answer 2

Use tac, so you dont have to read the whole file:

tac FILE |egrep -m 1 .

Answer 3

How about using grep to filter out the blank lines first?

$ cat rjh
1
2
3


$ grep "." rjh | tail -1
3

Answer 4

Instead of tac you can use tail -r if available.

tail -r | grep -m 1 '.'

Answer 5

if you want to omit any whitespaces, ie, spaces/tabs at the end of the line, not just empty lines

awk 'NF{p=$0}END{print p}' file

Answer 6

If tail -r isn't available and you don't have egrep, the following works nicely:

tac $FILE | grep -m 1 '.'

As you can see, it's a combination of two of the previous answers.

Answer 7

Print the last visually non-empty line:

tac my_file.txt | grep -vm 1 '^[[:space:]]*$'

or

awk 'BEGIN{RS="\r?\n[[:space:]]*"}END{print}' my_file.txt

The \r? is probably needed for DOS/Windows text files. Character classes such as [:space:] work with GNU versions of awk (i.e. gawk) and grep at least. With some other implementations you have to enumerate space characters manually in the code i.e. replace [:space:] by \t \n\f\r\v. I do not know what would be the proper way to deal with backspace characters (\b) in the file.

The following works only with [:space:]:

tac my_file.txt | grep -m 1 '[^[:space:]]'