Cfrtjryk

I know find was specified, but this sounds like a job for rsync.

I most often use the following:

rsync -axuv --delete-after --progress Source/ Target/

Here is a good example if you want to only move files of a particular file-type (example):

rsync -rv --include '*/' --include '*.js' --exclude '*' --prune-empty-dirs Source/ Target/

It's not as efficient, but the code is easier to read and understand, in my opinion, if you just copy the files and then delete afterwards.

find /original/file/path/* -mtime +7 -exec cp {} /new/file/path/ ;

find /original/file/path/* -mtime +7 -exec rm -rf {} ;

Notice: Flaw discovered by @MV for automated operations:

Using two separate operations is risky. If some files become older than 7 days
while the copy operation is done, they won't be copied but they will be deleted
by the delete operation. For something being done manually once this may not be
an issue, but for automated scripts this may lead to data loss

you could do it using two instances of find(1)

There's always cpio(1)

(cd "$soure" && find … | cpio -pdVmu "$destination")

Check the arguments for cpio. The ones I gave

You can do this by appending the absolute path of the file returned by find to your destination path:

find "$soure" -type f -mtime "-$days" -print0 | xargs -0 -I {} sh -c '

    file="{}"

    destination="'"$destination"'"

    mkdir -p "$destination/${file%/*}"

    mv "$file" "$destination/$file"'

Better (fastest & without consuming storage space by doing copy instead of move), also is not affected by the file-names if they contain special characters in their names:

export destination

find "$soure" -type f "-$days" -print0 | xargs -0 -n 10 bash -c '

for file in "$@"; do

  echo -n "Moving $file to $destination/"`dirname "$file"`" ... "

  mkdir -p "$destination"/`dirname "$file"`

  mv -f "$file" "$destination"/`dirname "$file"`/ || echo " fail !" && echo "done."

done'

Or faster, moving a bunch of files in the same time for multi-CPU, using the "parallel" command:

echo "Move oldest $days files from $soure to $destination in parallel (each 10 files by "`parallel --number-of-cores`" jobs):"

function move_files {

  for file in "$@"; do

    echo -n "Moving $file to $destination/"`dirname "$file"`" ... "

    mkdir -p "$destination"/`dirname "$file"`

    mv -f "$file" "$destination"/`dirname "$file"`/ || echo " fail !" && echo "done."

  done

}

export -f move_files

export destination

find "$soure" -type f "-$days" -print0 | parallel -0 -n 10 move_files

P.S.: You have a typo, "soure" should be "source". I kept the variable name.

Because there seem to be no really easy solution to this and I need it very often, I created this open source utility for linux (requires python): https://github.com/benapetr/smv

There are multiple ways how you could use it to achieve what you need but probably most simple would be something like this:

 # -vf = verbose + force (doesn't stop on errors)

smv -vf `find some_folder -type f -mtime "-$days"` target_folder

You can additionally run it in dry mode so that it doesn't do anything but print what it would do

smv -rvf `find some_folder -type f -mtime "-$days"` target_folder

Or in case that list of files is too long to fit into argument line and you don't mind executing python for every single file, then

find "$soure" -type f -mtime "-$days" -exec smv {} "$destination" ;

Hope this will help someone

This is less elegant but easy if the number / size of files isn't too great

Zip your files together into a zip archive, and then unzip at the destination without the -j option. By default, zip will create the relative directory structure.

Try this way:

IFS=$'n'

for f in `find "$soure" -type f -mtime "-$days"`;

do

  mkdir -p "$destination"/`dirname $f`;

  mv $f "$destination"/`dirname $f`;

done