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I have the following files.

root@VMBOX:/client# ls -lrt /client/*/ver* /client/*/*/ver* -rw-r--r-- 1 root root 0 Jul 5 18:54 /client/folder299/version_1 -rw-r--r-- 1 root root 0 Jul 5 18:58 /client/ifolder299/ifolder/version_a -rw-r--r-- 1 root root 0 Jul 5 18:58 /client/ifolder300/ifolder1/version_b -rw-r--r-- 1 root root 0 Jul 5 18:58 /client/ifolder301/ifolder2/version_c -rw-r--r-- 1 root root 0 Jul 5 18:58 /client/ifolder302/ifolder3/version_d -rw-r--r-- 1 root root 0 Jul 19 13:35 /client/folder300/version_2 -rw-r--r-- 1 root root 0 Jul 19 13:35 /client/folder301/version_3 -rw-r--r-- 1 root root 0 Jul 19 13:36 /client/folder302/version_4

I am trying to get the latest version file for a pattern matching an ID. Example is shown below.

root@VMBOX:/client# ls -lrt /client/*/ver* /client/*/*/ver* | grep 299 -rw-r--r-- 1 root root 0 Jul 5 18:54 /client/folder299/version_1 -rw-r--r-- 1 root root 0 Jul 5 18:58 /client/ifolder299/ifolder/version_a

The latest version is version_a in the above example.

root@VMBOX:/client# ls -lrt /client/*/ver* /client/*/*/ver* | grep 299 | tail -1 -rw-r--r-- 1 root root 0 Jul 5 18:58 /client/ifolder299/ifolder/version_a

I am told that this approach is not good to find a file (Why *not* parse `ls`? )and am looking for an alternative way like https://stackoverflow.com/a/26766782/9316558. Please let me know if something is not clear.

Update:

From below answer by Jasen, I could get the latest file in the path /client

find /client -path "*299*" -printf "%T@ %P\n" | sort -n | tail -1

But, the above command gives the latest file. I am looking for finding the latest version file.

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  • Will the latest version file always be the one that sorts last, or should file timestamps be used, or some other scheme? Commented Jul 19, 2018 at 8:55
  • @Kusalananda Yes, the latest version file will always be the one that sorts last. I mean the one that is created recently. Commented Jul 19, 2018 at 9:29
  • Sorry I should have been more specific. Sorts last, by name? ... taking into account that version_3 sorts after version_10. Commented Jul 19, 2018 at 9:30
  • @Raj: Does it have to be bash? If you would write your script in Zsh, you could do for instance a files=( /client/*299*/ver*(Nom) ) to get an array of files sorted in ascending order by modification time (o means sorting, m means modification time, N causes an empty array to be generated if no matching files exist). If you need descending order, use (^Nom) instead. Commented Jul 19, 2018 at 9:37
  • 1
    Please re-tag your Question to Solaris, as it's apparent from comments that you're only connecting from Linux, then running find on Solaris. Commented Jul 19, 2018 at 19:24

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you can combine find and sort

find -path "some pattern" -printf "%T@ %P\n" | sort -n | tail -1
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  • Sorry. Should I be running the command like this? find -name "299" -printf "%T@ %P\n" | sort -n | tail -1 Commented Jul 19, 2018 at 10:49
  • possibly -path "*299*" it understands the normal shell wildcards Commented Jul 19, 2018 at 10:50
  • it turns out -name was wrong and you need -path th match on directory names. answer edited. Commented Jul 19, 2018 at 10:54
  • yeah, you can add a start location before -path if you want search other than the current directory, see the man page for find - there's many other options. Commented Jul 19, 2018 at 11:16
  • Thanks for the update. The command seems to work if executed from inside the folder where I need this value from. root@RV-VMBOX:/client# find -path "*299*" -printf "%T@ %P\n" | sort -n | tail -1 1530797283.5560000000 ifolder299/ifolder/version_a In my case, I need to execute this command remotely as a user of the system. So, I understand that this will search in the home directory of the above user, where this folder doesn't exist. Is there a way in which we can extend the -path's argument to be more precise with the path? Commented Jul 19, 2018 at 11:18

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