3

I have recovered files and currently they have this structure:

root/MD5_of_file1/file1
root/MD5_of_file2/file2
...
root/MD5_of_filen/filen

Obviously, duplicates are now in the same folder. The filename contains no information, just the block number at which it was found during the recovery.

I want to flatten the structure, keeping only one file for each MD5. How could I do this efficiently ?

Just to be clear, here are some actual data:

feceee0fc150d191c5fd48ca6acee2f6
feceee0fc150d191c5fd48ca6acee2f6/f225407559.odt
feceee0fc150d191c5fd48ca6acee2f6/f94654911.odt
e905bb0a76c0055a2be1b8285d39c715
e905bb0a76c0055a2be1b8285d39c715/f0702423.odt
e905bb0a76c0055a2be1b8285d39c715/f26479232.odt
e905bb0a76c0055a2be1b8285d39c715/f3084695.odt

I want to flatten to something like this:

f225407559.odt
f0702423.odt

but there are no guarantees that filenames are distinct. Files could easily be renamed to the corresponding MD5 of their content, which is already computed, as it is the name of the folder in which they currently are.

  • ls root/MD5_of_file1/file* | grep -v 'root/MD5_of_file1/file1' | while read f ; do rm "$f" ; done – Jonah Nov 12 '15 at 16:12
  • @younes I'm pretty sure I've read something about not parsing the output of ls.. but thanks for your comment. – alecail Nov 12 '15 at 16:20
3
for i in *(/); do  mv $i/*([1]) $i.odt;  rm -rf $i; done

It's using zsh glob qualifier: *([1]) selects the first file in alphanumerical order

  • And just rm -rf $i, to ditch the directory altogether. – muru Nov 12 '15 at 17:33
  • That's better. A different approach (that should also work with other shells) would be something like this: for d in ./*/; do set -- "${d}"*; mv "$1" "${d%/*}.odt" && rm -rf "$d";done – don_crissti Nov 12 '15 at 21:25
0

In two steps:

perl-rename 's;/([^/]*)/[^/]*$;/\1_file;' foo/**/*
rmdir foo/**/

Example:

$ tree foo
foo
├── e905bb0a76c0055a2be1b8285d39c715
│   ├── f0702423.odt
│   ├── f26479232.odt
│   └── f3084695.odt
└── feceee0fc150d191c5fd48ca6acee2f6
    ├── f225407559.odt
    └── f94654911.odt

2 directories, 5 files
$ perl-rename -n 's;/([^/]*)/[^/]*$;/\1_file;' foo/**/*
foo/e905bb0a76c0055a2be1b8285d39c715/f0702423.odt -> foo/e905bb0a76c0055a2be1b8285d39c715_file
foo/e905bb0a76c0055a2be1b8285d39c715/f26479232.odt -> foo/e905bb0a76c0055a2be1b8285d39c715_file
foo/e905bb0a76c0055a2be1b8285d39c715/f3084695.odt -> foo/e905bb0a76c0055a2be1b8285d39c715_file
foo/feceee0fc150d191c5fd48ca6acee2f6/f225407559.odt -> foo/feceee0fc150d191c5fd48ca6acee2f6_file
foo/feceee0fc150d191c5fd48ca6acee2f6/f94654911.odt -> foo/feceee0fc150d191c5fd48ca6acee2f6_file
$ perl-rename 's;/([^/]*)/[^/]*$;/\1_file;' foo/**/*   
$ rmdir foo/**/
rmdir: failed to remove ‘foo/’: Directory not empty
$ tree foo
foo
├── e905bb0a76c0055a2be1b8285d39c715_file
└── feceee0fc150d191c5fd48ca6acee2f6_file

0 directories, 2 files

Another way, using find, sort and awk:

find foo -type f | 
  sort -k2,2 -u -t/ |
  awk -F/ -v OFS=/ '{path=$0; file=$NF; NF--; cmd = "cp " path " " $0 "_" file;  ; system(cmd); system("rm -r "$0)}'

Example:

$ find foo -type f | sort -k2,2 -u -t/ | awk -F/ -v OFS=/ '{path=$0; file=$NF; NF--; cmd = "cp " path " " $0 "_" file;  ; print cmd; print "rm -r "$0}'      
cp foo/e905bb0a76c0055a2be1b8285d39c715/f3084695.odt foo/e905bb0a76c0055a2be1b8285d39c715_f3084695.odt
rm -r foo/e905bb0a76c0055a2be1b8285d39c715
cp foo/feceee0fc150d191c5fd48ca6acee2f6/f225407559.odt foo/feceee0fc150d191c5fd48ca6acee2f6_f225407559.odt
rm -r foo/feceee0fc150d191c5fd48ca6acee2f6
$ find foo -type f | sort -k2,2 -u -t/ | awk -F/ -v OFS=/ '{path=$0; file=$NF; NF--; cmd = "cp " path " " $0 "_" file;  ; system(cmd); system("rm -r "$0)}'  
$ tree foo
foo
├── e905bb0a76c0055a2be1b8285d39c715_f3084695.odt
└── feceee0fc150d191c5fd48ca6acee2f6_f225407559.odt

0 directories, 2 files
  • So does the first step overwrites each file in each subfolder with its siblings ? I had in mind something that would pick one and drop the others, but if it get the job done.. – alecail Nov 12 '15 at 16:34
  • @AntoineLecaille overwrites each file. I can't think of a simple command to do it the other way. :( – muru Nov 12 '15 at 16:37
  • I'm on something with zsh glob qualifiers.. – alecail Nov 12 '15 at 16:44
  • @AntoineLecaille see update. – muru Nov 12 '15 at 16:56
  • 1
    I'd mv the file you keep, or at least cp -p, and awk can do the select-first: find foo -type f | awk -F/ -vOFS=/ '{path=$0;file=$NF;NF--} !already[$0]++ {system("mv "path" "$0"_"file"); system("rm -r "$0)}' – dave_thompson_085 Nov 12 '15 at 22:18

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