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Bash script - Remove the 3 top level of a full path filename

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# 1  
Old 11-28-2016
Bash script - Remove the 3 top level of a full path filename
Hello.

Source file are in : /a/b/c/d/e/f/g/some_file
Destination is : /d/e where sub-directories "f" and "g" may missing or not.

After copying I want /a/b/c/d/e/f/g/file1 in /d/e/f/g/file1
On source /a is top-level directory
On destination /d is top-level directory
I would like something like :
Code:
SRC="/a/b/c/d/e"
DEST="/d/e"

   cp_function --parents "$SRC"   "$DEST"

Any help is welcome.
# 2  
Old 11-28-2016
I'm not sure i follow ... but it looks like rsync could do the job for you ?
# 3  
Old 11-28-2016
Quote:
Originally Posted by Peasant
I'm not sure i follow ... but it looks like rsync could do the job for you ?
I have a try.

---------- Post updated at 16:46 ---------- Previous update was at 16:42 ----------

I have done this which works :
Code:
for MY_FILE in $( find "$SRC" -type f \( -iname "*$SEARCH1*" ! -iname "*~" \)  ) ; do
                    A_PATH=$( echo $MY_FILE | cut -d'/' -f6- )
                    DEST_PATH="/$A_PATH"
                    DEST_DIR_NEW=${DEST_PATH%/*}
                    mkdir -p "$DEST_DIR_NEW"
                    cp -fv  "$MY_FILE"  "$DEST_PATH"
done

---------- Post updated at 17:03 ---------- Previous update was at 16:46 ----------

Quote:
Originally Posted by Peasant
I'm not sure i follow ... but it looks like rsync could do the job for you ?
rsync does not create missing directories in destination.
So it is not better than cp
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