Anirudh's questions

I wanted to find the number of image files within a directory. I ended up using this :

find . -type f  -exec file {} \; | grep -c -i 'image'

This feels like an inefficient way of doing it. I could think of another way to do it - using an OR statement, but I couldn't figure out from the man page how this could be done.

Any ideas how this could be done more efficiently? I don't want to execute the commands above on a directory with a rather large number of files and have it take forever.

I have seen Cron sync mysql tables . But I cannot use replication. I used to use percona-toolkit and it worked perfectly. I could run the sync command on demand or just run it via cron. It would compare checksums on the two tables and do inserts, updates, deletes etc. However Perl(DBD::mysql) and MySQL have some incompatibilities in a new server I'm going to run this on and I cannot use pt-table-sync. Is there a similar solution which uses something other than Perl/DBD?

Edit : (more details, clarity)

1. Both the source and target tables are live tables and in constant use. Hence a solution with the table not existing (eg. a DROP TABLE is done) would not be acceptable.
2. Cannot use replication or any such modifications to the server itself. The solution has to work on the client side.
3. In this particular scenario both the tables are in the same server, but different DBs (eg. db1.tbl db2.tbl). However, a solution which does not rely on this fact would definitely be a bonus
4. Network latency is unlikely to be an issue. In this case the script runs on a server in the same datacenter.
5. Cannot use Perl (incompatibility between Perl and MySQL - 64 bit vs 32 bit)