-1
#!/bin/bash
echo "hi"
path="/home/alert/VideoApplicationAPI.v1/logs"
dayDiff=365

DATE=`date +%Y-%m-%d`
for filename in $path/*.*; do
    modDate=$(stat -c %y  "$filename") 
    #modDate=$(date -r   "$filename"+%s) 
    modDate=${modDate%%  *}
    echo $filename:$modDate

    #lastUpdate=$(stat -c %y "$filename")
    now="$(date +%s)"
    diff="${now}-${lastUpdate}"
done

echo $DATE 
1

This is close to a repost of this question. Adapting their script to your format gives us this set of commands:

echo "$(( $(date -d "$d2" +%s) - $(date -d "$d1" +%s) )) / 86400" | bc -l

where $d1 is the smaller (earlier date) and $d2 is the larger (later) date.

So, as far as I can tell, this should do the trick:

echo "$(( $(date -d "$modDate" +%s) - $(date -d "$now" +%s) )) / 86400" | bc -l

To clarify,

bc -l

Is, according to its manpage,

...a language that supports arbitrary precision numbers with interactive execution of statements. There are some similarities in the syntax to the C programming language.

It allows you to get decimal numbers in your answer, as most shells only support integer division.

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0

I'm not sure what you're trying to modify here, as far as I know, stat -c only works on GNU stat, and stat -c %y gives the output in the format 2019-03-14 14:21:32.704211521 +0200. You're trying to remove anything after a double space, but there isn't one.


In any case, if you want to do arithmetic on the timestamp, it's better to just take it as seconds since the Epoch, i.e. -c %Y, not -c %y. Then, to get the difference (in seconds), you can use arithmetic expansion $(( .. )) in the shell:

$ diff=$(( "$(date +%s)" - "$(stat -c %Y "$file")" ))
$ echo $diff
86527

To get that that as hours, minutes and seconds, just make the appropriate divisions and take the remainders:

$ s=$(( diff % 60 )); m=$(( diff / 60 % 60 )); h=$(( diff / 3600 ))
$ printf "%d:%02d:%02d\n" "$h" "$m" "$s"
24:02:07

Or use e.g. bc to get the time in fractional days, as @Pheric's answer shows.

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