How to find java runtime in dos on Windows 7

I have been unsuccessful in finding the right DOS commands to achieve this and would appreciate some help.

This is an extract from my registry:

[HKEY_LOCAL_MACHINE\SOFTWARE\JavaSoft\Java Runtime Environment\1.6]
"JavaHome"="G:\\Program Files (x86)\\Java\\jre6"

Not sure why it is showing the double backslash...

I've tried various commands (in BATch files), such as the following, without success:

- - - -
@echo off
set KeyName=SOFTWARE\JavaSoft\Java Runtime Environment\1.6
REM set KeyName=HKEY_LOCAL_MACHINE\SOFTWARE\JavaSoft\Java Runtime Environment
set Cmd=reg query "%KeyName%" /s
for /f "tokens=2*" %%i in ('%Cmd% ^| find "JavaHome"') do set JAVA_HOME=%%j

set JAVA_HOME
pause
- - - -

Any help will be appreciated.

I know this can be set manually, but this is not what is needed.

Thank you for responding. I should have mentioned that your suggestion was one of the variations I had tried - sorry.

Here is a screen print showing the result of two different tests.



Thanks again for trying to help :)

Neutron16's change worked for me. Should you be using HKLM64? I only have a 32bit VM to test in right now.

Code:

cd\
set KeyName=HKLM\SOFTWARE\JavaSoft\Java Runtime Environment\1.6
set Cmd=reg query "%KeyName%" /s
for /f "tokens=2*" %%i in ('%Cmd% ^| find "JavaHome"') do set JAVA_HOME=%%j
set JAVA_HOME
pause

Excellent, thank you.

The successful search string is: HKEY_LOCAL_MACHINE\Software\Wow6432Node\JavaSoft\Java Runtime Environment\1.6

Great