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I have a requirement to identify sequence gap in a set of files. Sequence starts at FILENAME_0001 and ends at FILENAME_9999. After this the sequence is restarted from 0001.

To implement a proper sequence check I used ls -rt to pick the files in order of modified time and the compared with the previous files sequence number. If the previous file was 9999 I check whether the next one is 0001 (to accommodate the sequence reset).

Recently I came across a scenario where files were listed in the below order:

FILENAME_0001 
FILENAME_0002
FILENAME_0005
FILENAME_0003
FILENAME_0004
FILENAME_0006
FILENAME_0007

This was because files 3, 4 & 5 had the same modified time to the second. Only the millisecond was different. So I am guessing ls -rt considers only upto the seconds. Could someone suggest a workaround?

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  • If I understand correctly, you're creating the files ordered by the sequence no (which is also increasing time), but not all files are created - so there are missing values in the sequence, which you're trying to identify. If so, cant you just sort the files by their seq nos and look for the missing ones ? Commented Feb 11, 2016 at 10:08
  • Yes. But note that seq numbers reset at 9999. So the way you suggested it will fail when I have files FILENAME_9999, FILENAME_0001, FILENAME_0002 Commented Feb 11, 2016 at 10:19
  • What Unix variant are you running this on? Getting at sub-second timestamps isn't portable. Commented Feb 11, 2016 at 22:24

2 Answers 2

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If your find has printf, print out the mtime in seconds followed by the filename, then use sort, and finally cut:

find . -type f -printf "%T@\t%f\n" |
sort -k 1n -k 2 |
cut -f 2-

The find outputs TIMESTAMP FILENAME on each line. The sort first sorts the timestamps in numerical order. If the timestamps are equal, it will use the filename as a last resort. The cut removes the timestamp from the output.

EDIT: Your perl solution works, but I would do it differently. Here's the simplest:

find . -type f -print | 
perl -lne 'print (((stat($_))[9]."\t".$_)' |
sort -k 1n -k 2 |
cut -f 2-

No need to convert the time to a string and back again. Just output stat's mtime as a numeric value as find would have done.

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  • Sounds good. I'll test it tomorrow and let you know. I don't have access to the system now. Commented Feb 11, 2016 at 12:11
  • This work in my home. But the actual box where I needed it to run doesn't support it 😔. The version of find I have installed doesn't up port printf Commented Feb 12, 2016 at 2:59
  • What kind of system is it? (uname -a should be a clue.) Does it support the stat command with format control? (stat -c ) Can you install GNU find? Commented Feb 12, 2016 at 7:00
  • I use HP-UX system. However I am marking your answer also as correct since if worked on my home machine. Commented Feb 12, 2016 at 7:07
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Finally got it working. I used the below code:

for FILENAME in $(ls...); do
FILE_TIME=$(perl -e '@d=localtime ((stat(shift))[9]); printf "%4d%02d%02d%02d%02d%02d\n", $d[5]+1900,$d[4]+1,$d[3],$d[2],$d[1],$d[0]' $FILENAME)
echo "$FILE_TIME $FILENAME"
done | sort -k 1n -k 2 | cut -d" " -f2

I use HP-UX system.

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  • This works, but for a slightly less cumbersome way, check out my edit to my answer :) Commented Feb 12, 2016 at 11:13

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