LinuxQuestions.org - rsync - clever suggestion needed

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rsync - clever suggestion needed

Hi,

Start off point:
Server1 has file1 at /home/user1/
Server2 has file1 at /home/user2/

How do I use rsync to delete file1 at server2, on the basis that it exists on server1?

I think I need the inverse of --delete, which deletes everything that isn't at server1, but is at server2.
Does rsync have a function that will remove files from the destination that are already at the source?

Cheers,
Ape

You could build a list of all files you have in the source, put that in an exclude file and call rsync with --delete-exclude.

But rsync seems not to be the correct tool as it focuses on syncing, and what you want is explicit non-syncing. It might be better to build a list of files you have in source, open a SSH to dest and delete files you have in dest. If the file doesn't exist, you won't delete it.

jlinkels

Non-rsync solution

Hi,

I've solved this by:

Code:

LOCAL_DIR=/home/user2/

REMOTE_DIR=/home/user1/

ACCOUNT=user1

HOST=Server1

#Create a list of files at the remote location:

REMOTE_DIR_LIST=`ssh ${ACCOUNT}@${HOST} "find ${REMOTE_DIR} -name \"file*\" -type f -printf \"%f \" | sed 's/\s$//'"`

#Create a list of files at the local location:

LOCAL_DIR_LIST=`find ${LOCAL_DIR} -name "file*" -type f -printf "%f " | sed 's/\s$//'`

for ITEM in ${LOCAL_DIR_LIST}; do

        if [[ ${REMOTE_DIR_LIST} =~ ${ITEM} ]]; then

                REMOVAL_LIST=${ITEM}" "${REMOVAL_LIST}

        fi

done

REMOVAL_LIST=`echo ${REMOVAL_LIST} | sed 's/\s$//'`

if [[ ${REMOVAL_LIST} != "" ]]; then

        files_removed_OK=0

        files_removed_NOK=0

        for file in ${REMOVAL_LIST}; do

                if [[ `rm -vf ${LOCAL_DIR}/${file}` ]]; then

                        (( files_removed_OK++ ))

                else

                        (( files_removed_NOK++ ))

                fi

        done

fi