Shell Programming and Scripting

Hello

I have three directory structures for code releases.

Each directory structure looks like this:

bash-3.00$ ls -R | more

.:
Test_Release_1
Test_Release_2
Test_Release_3

./Test_Release_1/dbcode:

rp_online_import_srdp.pkb-1
srdp_ar_validation.pkb-1
srdp_rp_aclaim_detail_qry_pkg.pkb-1
srdp_rp_common_qry_pkg.pkb-1

./Test_Release_2/dbcode:

rp_online_import_srdp.pkb-2
srdp_ar_validation.pkb-2
srdp_rp_aclaim_detail_qry_pkg.pkb-2
srdp_rp_common_qry_pkg.pkb-2

./Test_Release_3/dbcode:

rp_online_import_srdp.pkb-3
srdp_ar_validation.pkb-3
srdp_rp_aclaim_detail_qry_pkg.pkb-3
srdp_rp_common_qry_pkg.pkb-3

What I'm wanting to do is, is to create a composite directory called "Test_Releases_1-3" which will have the same structure (with a dbcode directory), but will only copy scripts with the highest version number into the dbcode folder of the new composite directory "Test_Releases_1-3". In the new composite release structure I only want to get the very latest versions of scripts. The code release directories often contain multiple versions of the same package.

I'm needing to write a script which will create the composite release directory automatically, only extracting the latest version of packages from the original release directories.

Hope that made some sort of sense.

Thanks in advance
Glyn

I am not clear with the requirement.
what are the version number? are those which are mentioned at the end of the file names?

It will be better if you list the files from the above example which you want to copy in the new directory.