Sorting LS Output by Time – Handling Too Many Files

lsshell

I know that I can sort the output of ls by time with the -t option.

I know that when I have so many files that they don't fit in a single ls invocation, I can normally use xargs (or find ... -exec ... {} +) to let ls be called multiple times.

How can I combine the two? I have more files than fit on the command-line, and wish to list them sorted by time. find . -type f -exec ls -t {} + doesn't work, because supposing exactly 1000 file names fit on the command-line, and 3000 files are present, this will run ls -t [first 1000 files]; ls -t [second 1000 files]; ls -t [last 1000 files], where the last 1000 files find sees may well have a modification time before any of the first 1000. It doesn't seem like anything involving xargs or equivalent has any chance whatsoever of working, it seems like that approach is fundamentally flawed, but I cannot find a way that does work.

Best Answer

ls -t on its own will list all files in the current directory with that sorting, without ever needing to list them on the command line at all.

If you need the recursion behaviour of find, or to do some other tests on the files, you can have find generate timestamped entries, either through stat or through GNU find's -printf option and sort it. Something like:

find . -type f -printf '%T@ %p\0' | sort -zn

-printf '%T@ %p\0' generates null-separated Unix timestamp (%T@)-filename (%p) pairs. sort -z is also a non-standard GNU extension, which uses null-delimited records to be filename-safe. The sort option is supported in most of the BSDs too, but -printf is GNU-only as far as I know.

You can cut that output back into filenames only, or any other format you like.

Related Solutions

Bash – Highlight the three last updated files in ls output

The following seems to work for me

 grep --color -E -- "$(ls -rtl | tail -n3)|$" <(ls -l)

It uses grep with highlight on input ls -l and uses a regular expression to search for either of the inputs for the three oldest command. It also search for the end-of-line $ in order to print the whole file.

You can also put it in a function, such that you can use lll * with multiple arguments, just as you would use ls

function lll ()
{
    command grep --color -E -- "$(ls -rtl $@ | tail -n3)|$" <(ls -l $@)
}

Shell Find – Locate Directories Without Subdirectories

You don't need to use awk at all. Use the built-in tests that ksh provides, something like this:

#!/bin/ksh

for NAME in *
do
    FOUND=no
    if [[ -d $NAME && $NAME != '.' && $NAME != '..' ]]
    then
        for SUBNAME in $NAME/*
        do
            if [[ -d $SUBNAME ]]
            then
                FOUND=yes
                break
            fi
        done
        if [[ $FOUND == no ]]
        then
            echo Found only files in $NAME
        fi
    fi
done

That little script looks in all the directories in the current directory, and tells you if they only contain files, no sub-directories.

Best Answer

Related Solutions

Bash – Highlight the three last updated files in ls output

Shell Find – Locate Directories Without Subdirectories

Related Question