Determine if JRE version is compliant through Batch Script

Question

I am deploying multiple programs to remote systems. Some of the programs require JRE 1.41 or later to run. I would like to create a bat script to not only identify the JRE version, but identify if the client has a newer version of JRE. So far I have been able to identify the latest version. However, I do not know how to compare to a later version of JRE. For example lets say JRE version 6.0 comes out. At this point the program will probably be compliant with my machine, but I don't know how to make my script identify later versions, since batch files only compare and don't have many math features. I thought about doing a comparison between every java version that has been deployed.(I haven't even found a comprehensive list either) But this may not provide support for newer versions of JRE Any suggestions? Below is what I have so far.
@echo off
start /w regedit /e reg1.txt "HKEY_LOCAL_MACHINE\SOFTWARE\JavaSoft\Java Runtime Environment"
type reg1.txt | find "CurrentVersion" > reg2.txt
for /f "tokens=2 delims==" %%x in (reg2.txt) do set JavaTemp=%%~x
if %JavaTemp% == 1.41 echo G2G
if %JavaTemp% == 1.3 echo Failed. You need to upgrade JRE to at least 1.41

Answer 1

does anyone know how (or if it's possible) to
determine if a JRE is the international version?
thanks.There is no API to provide this answer conclusively. However, one might guess.
For example, in the non-localized JRE, you will not have the Japanese locale support. So you could do something like this:
Locale[] supportedLocales = DateFormat.getAvailableLocales();
If the Locale array contains a Locale for ja_JP, then you have the international version. It's simple, but perhaps not the cleanest way of determining the answer to your question.
Regards,
John O'Conner
http://www.joconner.com/javai18n/