How to Sort Files by Modification Time

findlssorttimestampswildcards

I've seen many questions and answers here using a construction along the lines of

list_dir=`ls -t /path/to/dir/`
for i in $list_dir; do

ls -t | while read i; do

Now, I know that you shouldn't use ls in scripts because it breaks easily; but I can't find a better way of operating on files in order from last-modified to most-recently-modified (or vice versa).

I can use something like:

find . -type f -printf '%T@ %p\n' | sort -n | cut -d ' ' -f 2- | while read i; do...

…but this will still break with any files that have newlines in their names, and is much uglier to boot. Is there a better way?

Best Answer

"Don't use ls in scripts" is a problem with POSIX ls "only"; for GNU ls see --quoting-style=.

GNU sort solves the problem with --zero-terminated.

If it must be compatible then you could use find ... -exec for passing one file name at a time to a script which does the escaping. If at least bash is available:

start cmd:> testfunc () { echo "${1//$'\n'/\n}"; }
start cmd:> testfunc a$'\n'b
a\nb

Related Solutions

Shell – list all files newer than given timestamp and sort them

find supports a lot of date input formats. The simplest format to obtain is YYYYMMDD HH:MM:SS. You already have the digits in the right order, all you have to do is extract the first group (${timestamp%??????}: take all but the last 6 characters; ${timestamp#????????}: take all but the first 8 characters), and keep going, appending punctuation then the next group as you go along.

timestamp=20130207003851
timestring=${timestamp%??????}; timestamp=${timestamp#????????}
timestring="$timestring ${timestamp%????}"; timestamp=${timestamp#??}
timestring="$timestring:${timestamp%??}:${timestamp#??}"

In bash (and ksh and zsh), but not in ash, you can use the more readable ${STRING_VARIABLE:OFFSET:LENGTH} construct.

timestring="${timestamp:0:8} ${timestamp:8:2}:${timestamp:10:2}:${timestamp:12:2}"

To sort files by date, print out the file names preceded by the dates and sort that, then strip the date prefix. Use -printf to control the output format. %TX prints a part of the modification time determined by X; if X is @, you get the number of seconds since the Unix epoch. Below I print three tab-separated columns: the time in sortable format, the file name, and the time in human-readable format; cut -f 2- removes the first column and the call to expand replaces the tab by enough spaces to accommodate all expected file names (adjust 40 as desired). This code assumes you have no newlines or tabs in file names.

find -maxdepth 1 -type f \
     -newermt "$timestring" -printf '%T@\t%f\t%Tb %Td %TH:%TM\n' |
sort -k1n |
cut -f 2- |
expand -t 40

Bash – How to sort results from ls command by modification date (directories first)

what about something like this:

ls -ltr --group-directories-first

Best Answer

Related Solutions

Shell – list all files newer than given timestamp and sort them

Bash – How to sort results from ls command by modification date (directories first)

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