How to find newest file with given name?

This post will probably be boring for you, but this is mostly just a reminder to myself, written in form of a blog post.

So, I have a directory structure: /some/path/imported/DATE/TIME/file, where DATE is date of importing, in format YYYY-MM-DD, and TIME is time of importing, in format HHMMSS.

So, example paths look like this:

./2009-02-26/143251/5a6d001b94e47960fe41a262f70ed96a
./2009-02-26/143321/8e45f68421dad6129914fe068dfa5748
./2009-02-26/143407/aa04aa9c1e8f87b25fef98bd9a64e94d
./2009-02-26/143415/65180d1328e21959229e47b9288b6996
./2009-02-27/083542/5a6d001b94e47960fe41a262f70ed96a
./2009-02-27/084906/aa04aa9c1e8f87b25fef98bd9a64e94d
./2009-02-27/084926/65180d1328e21959229e47b9288b6996
./2009-02-27/155648/65180d1328e21959229e47b9288b6996

As you can see some of the files were imported many times.

Now, I need to find the latest import of given file.

So, I need a way to convert above list into:

./2009-02-26/143321/8e45f68421dad6129914fe068dfa5748
./2009-02-27/083542/5a6d001b94e47960fe41a262f70ed96a
./2009-02-27/084906/aa04aa9c1e8f87b25fef98bd9a64e94d
./2009-02-27/155648/65180d1328e21959229e47b9288b6996

Of course – with 10 imports, it's simple. But what if I had 10000 of them?

Luckily, it is rather simple:

find . -mindepth 3 -maxdepth 3 -exec basename {} \; | \
    sort -u | \
    while read DIR; \
    do \
        find . -name "$DIR" | \
        sort | \
        tail -n 1; \
    done

Of course I typed it originally as one-liner 🙂

While writing the post I realized I could do better:

find . -mindepth 3 -maxdepth 3 | \
    sort -r -t/ -k4,4 -k2,2 | \
    awk -F/ 'BEGIN{prev="/"} ($4!=prev) {print $0; prev=$4}'

Well. I understand the code, and what it does, but it doesn't change the fact that I'm not really fan of shell programming.