cat last 5 files in a dir

Question

I have a directory with filenames of the format $date.txt. I would like to cat 5 latest files from it. Is there a more elegant solution to that, than

for f in 2*.txt; do echo $f; done | tail -5 | while read f; do cat $f; done

Answer 1

In ksh93, bash or zsh:

files=( 2*.txt )
cat "${files[@]: -5}"

This would create an array of the filenames matching the pattern 2*.txt. It would then output the contents of the last five of these.

In zsh, you can also specify a range of files as part of its glob qualifiers:

cat 2*.txt([-5,-1])

In any POSIX shell, this may also be done through

set -- 2*.txt
while [ "$#" -gt 5 ]; do shift; done   # or: [ "$#" -gt 5 ] && shift "$(( $# - 5 ))"
cat "$@"

This sets the positional parameters to the filenames matching the pattern. It then shifts off the names from the beginning of the list until the list only has five elements in it. cat is then invoked on the remaining filenames.

In all of these solutions, the files would be sorted lexicographically. Filenames with spaces or newlines are handled correctly.

Answer 2

I would use a while read loop, this would handle files with spaces.

ls 2*.txt | tail -5 | while read loop
do
  cat "$loop" 
done 

If you always want the 5 most recent you could change it to ls -tr | tail -5

Answer 3

Based on the question, with no prospect of spaces and 2*.txt happily returning correct ordering, how about

cat $(ls 2*.txt|tail -5)

Answer 4

As mentioned latest in question, we can check with the modification time of files in the directory.

 cat $(ls -ltr *.txt | tail -n5 | awk '{print $NF}' | tr '\n' ' ')

• ls -ltr - will list files with respect to modification timestamp (reverse)

• tail -n5 - will take last 5 files

• $NF - last field

• tr - transpose column to row