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B.Tanner
Asked: 2021-01-21 04:37:55 +0800 CST

`ls -l` output duration instead of date/time eg "5 mins"/"6 hours"/"7 days" etc?

  • 772

If I last modified a file 5 minutes ago, is it possible to make ls-l output something like "5 mins" instead of the actual date/time?

command-line

4 Answers

  1. I use stat to get metadata info on the files. Some examples:

    stat -c $'%y\t%n' * | sort -n

    Output looks like this:

    2020-01-27 11:52:25.681249958 +0200 CHANGELOG.md

    Then to lookup a single file

    stat CHANGELOG.md

    and output looks like this:

      File: CHANGELOG.md
  Size: 94          Blocks: 8          IO Block: 4096   regular file
  Device: fd00h/64768d  Inode: 6029378     Links: 1
  Access: (0644/-rw-r--r--)  Uid: (  998/example)   Gid: (  998/example)
  Access: 2020-11-12 17:47:34.768793021 +0200
  Modify: 2020-01-27 11:52:25.681249958 +0200
  Change: 2020-11-12 17:47:02.282093672 +0200
  Birth: -

    Otherwise you might need a small bash script to show you the difference between when the file was created and current time.

     LastUpdate="$(stat -c %Y myfile)"
 now="$(date +%s)"
 let diff="${now}-${lastUpdate}"
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  2. This is a continuation of an earlier script used for parsing
    ls -l with some enhancement:

    #!/bin/bash

while read -r p c u g s e n; do
    [[ $p = total ]] && { \
        echo "$p $c" 1>&2; continue; \
    }
    x=$(($EPOCHSECONDS - e))
    y=$((x / 60))
    z=$((x % 60))
    printf '%s\t%s\t%s\t%s\t%s\t%s\t%s\t%s\n' \
            "$p" \
            "$c" \
            "$u" \
            "$g" \
            "$s" \
            "$y minutes" \
            "$z seconds" \
            "$n"
done < <( \
    ls -LApl --color=force \
             --time-style=+%s \
             --quoting-style=shell-escape "$@" \
) \
| column -t -R 2,5,6,7 -s $'\t' -o ' '
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  3. One way:

    You can make ls print modification time in seconds since epoch, like so:

    ls -l --time-style="+%s"

    Then pipe the ls output to awk to do the logic, like so:

    ls -l --time-style="+%s" | awk -v now=$(date +%s) '{$6 = now - $6; d=int($6/60/60/24); h=int($6/60/60%24); m=int($6/60%60); s=int($6%60);$6 = "="d" Days "h" Hours "m" Mins "s" Secs ago="; print}' | column -t -s'='

    Because the command line is long, here is a break-up which is basically the same but can be visually more contained:

    ls -l --time-style="+%s" | \
awk -v now=$(date +%s) '{\
$6 = now - $6; \
d=int($6/60/60/24); \
h=int($6/60/60%24); \
m=int($6/60%60); \
s=int($6%60);\
$6 = "="d" Days "h" Hours "m" Mins "s" Secs ago="; print}' | column -t -s'='

    • -v now=$(date +%s) will assign the current time in seconds since epoch to now

    • $6 = now - $6 will calculate the time in seconds since file last modified.

    • d=int($6/60/60/24) will calculate the days.

    • h=int($6/60/60%24) will calculate remaining hours less than one day.

    • m=int($6/60%60) will calculate remaining minutes less than one hour.

    • s=int($6%60) will calculate remaining seconds less than one minute.

    • $6 = d" Days "h" Hours "m" Mins "s" Secs ago" will change the modification date to something like 2 Days 5 Hours 3 Mins 47 Secs ago.

    Another way:

    You can construct your own custom ls -l alternative in a bash script, like so:

    #!/bin/bash

for f in *
        do
        p=$(stat -c "%A=%h=%U=%G=%s" "$f")
        m=$(date -r "$f" "+%s")
        n=$(date "+%s")
        l=$((n-m))
        d=$((l/60/60/24))
        h=$((l/60/60%24))
        m=$((l/60%60))
        s=$((l%60))
        echo "$p= = =$d Days=$h Hours=$m Mins=$s=Secs= = =$f" | column -t -s'='
        done

    or make the output simpler and less time strict, like so:

    #!/bin/bash

for f in *
        do
        p=$(stat -c "%A %h %U %G %s" "$f")
        m=$(date -r "$f" "+%s")
        n=$(date "+%s")
        l=$((n-m))
        d=$((l/60/60/24))
        h=$((l/60/60%24))
        m=$((l/60%60))
        s=$((l%60))
        if (( $d >= 1 ))
                then
                t="$d Days"
        elif (( $h >= 1 ))
                then
                t="$h Hours"
        elif (( $m >= 1 ))
                then
                t="$m Mins"
        else
                t="$s Secs"
        fi 
        echo "$p= = =$t= = =$f" | column -t -s'='
        done

    Notice on usage:

    If you save either the above command or the above script in a file in your home directory, name the file lsl.sh and make it executable like so:

    chmod +x ~/lsl.sh

    you can add an alias like this:

    alias lsl='bash ~/lsl.sh'

    to your ~/.bashrc file so you can afterwords run the short command: `

    lsl

    from any directory and get the desired result.

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  4. Best Answer

    Answering my own question here because this is what I actually ended up doing.

    I could not get any satisfactory results using anything that parsed the output of ls -l into a word stream (bash, awk, column etc) because a) filenames with spaces in messed up, and b) ls -l's aligned columns messed up (for example, right-aligned filesizes). ls -l | column -t shows both problems at once.

    I will get shot down for the inefficiency in this, but this is my eventual script that uses sed to replace parts of the output of ls -l maintaining column width:

    day0=`date                     +"%b %_d"`
day1=`date --date "1 day  ago" +"%b %_d"`
day2=`date --date "2 days ago" +"%b %_d"`
day3=`date --date "3 days ago" +"%b %_d"`
day4=`date --date "4 days ago" +"%b %_d"`
day5=`date --date "5 days ago" +"%b %_d"`
day6=`date --date "6 days ago" +"%b %_d"`

min0=`date                        +"%H:%M"`
min1=`date --date "1 minute  ago" +"%H:%M"`
min2=`date --date "2 minutes ago" +"%H:%M"`
min3=`date --date "3 minutes ago" +"%H:%M"`
min4=`date --date "4 minutes ago" +"%H:%M"`
min5=`date --date "5 minutes ago" +"%H:%M"`
min6=`date --date "6 minutes ago" +"%H:%M"`
min7=`date --date "7 minutes ago" +"%H:%M"`
min8=`date --date "8 minutes ago" +"%H:%M"`
min9=`date --date "9 minutes ago" +"%H:%M"`

ls -lh --color=force --quoting-style=shell-escape "$@"|sed "\
  /$day6 \(.[0-9]:[0-9][0-9]\)/ s/$day6/6 days/; t
  /$day5 \(.[0-9]:[0-9][0-9]\)/ s/$day5/5 days/; t
  /$day4 \(.[0-9]:[0-9][0-9]\)/ s/$day4/4 days/; t
  /$day3 \(.[0-9]:[0-9][0-9]\)/ s/$day3/3 days/; t
  /$day2 \(.[0-9]:[0-9][0-9]\)/ s/$day2/2 days/; t
  /$day1 \(.[0-9]:[0-9][0-9]\)/ s/$day1/yesday/; t
  /$day0 \(.[0-9]:[0-9][0-9]\)/ s/$day0/ today/; T
  s/$min0/  now/; t
  s/$min1/1 min/; t
  s/$min2/2 min/; t
  s/$min3/3 min/; t
  s/$min4/4 min/; t
  s/$min5/5 min/; t
  s/$min6/6 min/; t
  s/$min7/7 min/; t
  s/$min8/8 min/; t
  s/$min9/9 min/; t
  "

    Sample output of /var/log/syslog*:

    -rw-r----- 1 syslog adm    9240  today   now /var/log/syslog
-rw-r----- 1 syslog adm 1279965  today 8 min /var/log/syslog.1
-rw-r----- 1 syslog adm   51750 2 days 14:12 /var/log/syslog.2.gz
-rw-r----- 1 syslog adm   14768 3 days 10:59 /var/log/syslog.3.gz
-rw-r----- 1 syslog adm    7767 4 days 10:04 /var/log/syslog.4.gz
-rw-r----- 1 syslog adm  119295 5 days 09:49 /var/log/syslog.5.gz
-rw-r----- 1 syslog adm   33450 6 days 13:06 /var/log/syslog.6.gz
-rw-r----- 1 syslog adm   21372 Jan 25 11:12 /var/log/syslog.7.gz

    (not sure why log appears here in red; it doesn't on my terminal.)

    This is only used from the command line so all those date commands don't matter, in fact I don't notice any delay at all.

    Update:

    time output for `ls -l /var/log/syslog*':

    real    0m0.002s
user    0m0.002s
sys     0m0.000s

    time output for for my new script:

    real    0m0.019s
user    0m0.018s
sys     0m0.003s
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