Shell Programming and Scripting

Hi,
using awk command I want to print filenames and the last line of each file, in a single command line statement.
I want to use 'awk', because I want to add more functionality to this logic later.

I tried the following on *.sh files in the current directory

Code:

find . -type f -name "*.sh" -exec awk 'NR = FNR {print FILENAME NR}' {} \;

but the above command prints all line numbers of every file and not just the last line
I understand that 'NR' stands for line number and not the line content itself

How can I print the filename, along with ONLY the last line (content) of the file?

Thanks,
-sri

You could use something like this:

Code:

find . -type f -name "*.sh" -exec awk '{s=$0};END{print FILENAME,s}' {} \;

Thanks Subbeh, it works !
Now, how can I print the filename and the last line of only those files where the last line is not like '' (empty) ?

In other words, I want list all filenames together with their last line where the file doesn't end with a new line

Thanks
-sri

You can simply use this:

Code:

find . -type f -name "*.sh" -exec awk '{s=$0};END{if(s)print FILENAME,s}' {} \;

... or simply

Code:

 find . -type f -name "*.sh" -exec awk 'END{if ($0) print FILENAME,$0}' {} \;

AWK is not required to preserve the value of the field variables ($0, $1, etc...) for the END section and some implementations don't. Subbeh's approach is the safest, most portable solution.

Regards,
Alister

Works with the original awk, mawk, nawk, gawk and gawk in its traditionnal mode. If OP needs to process big files, it's about two times faster.

See also http://awk.info/?awk1line