compare files in two directory on two servers using unix

Question

There is a similar question: compare files in two directory on remote server using unix

But my situation is I can not login the remote server, it's customer machine, so can not use rsync, but I can request customer to export the directory structure, tell me the dir/file's name, size, md5 value etc.

I only want to display the file name that in different content.

How to do it? best using ksh...

Thanks in advance.

Answer 1

It kind of depends whether file dates and times are important or file contents...

If the dates are important (e.g. it's a syncing thing you are investigating) I would be running find with ls -l.

If the contents are important, I would want to be checking MD5 checksums like this:

find . -type f -print -exec md5 -q {} \;   > somefile

That gives output like this:

./West Wales 14-Oct-09.axe
2c0c390bfc4206b8b88e11d537eacda8
./wl
44f84a91a98da15381a198e29417170c
./YOURFILE
ea102bc16e2b449e4ac6770b73cb9c50
./YOURFILE.BAK
a6ba1946cb666cb3b88ac31e6fb3f3f0
./z.html
b4554a1044abe07fd23d4580dd3055cc

Then on your local machine, read the file and calculate its checksum locally and compare that with the remote one:

#!/bin/bash
while read fname
do
    read remotemd5
    localmd5=$(md5 -q "$fname")
    if [ $remotemd5 != $localmd5 ]; then
       echo $fname $localmd5 $remotemd5 
    fi
done < file

Answer 2

Ask the customer to run a command to get all the relevant info, eg:

find . -type f -exec ls -l {} \; | sort > remote_dir.txt

Then run the same command locally:

find . -type f -exec ls -l {} \; | sort > local_dir.txt

Then you can use comm to compare the two and see what has changed.

comm -3 remote_dir.txt local_dir.txt