Question

Like the title says, how can I filter with grep (or similar bash tool) the line-before-the-last-line of a (variable length) file?

That is, show everything EXCEPT the penultimate line.

Thanks

Answer 1

You can use a combination of head and tail like this for example:

$ cat input 
one
two
three
HIDDEN
four
$ head -n -2 input ; tail -n 1 input 
one
two
three
four

From the coreutils head documentation:

‘-n k’
‘–lines=k’
Output the first k lines. However, if k starts with a ‘-’, print all but the last k lines of each file. Size multiplier suffixes are the same as with the -c option.

So the head -n -2 part strips all but the last two lines of its input.

This is unfortunately not portable. (POSIX does not allow negative values in the -n parameter.)