Win 7上的32位Java

Question

How do I make a 32-bit java 7 as the default in a 64-bit Windows 7?
When I type in java at the command prompt - it seems to look to a 64-bit java(which I have uninstalled) by default:

could not open "c:\program files\java\jre7\lib\amd64\jvm.cfg'

I am not able to locate how or why is it looking for this specific path.
The JAVA_HOME and PATH contain no reference to it.

How do I make it search for the 32-bit JDK that has been referred to in JAVA_HOME instead?

Answer 1

Please follow this tutorial for setting the PATH and JAVA_HOME on Windows 7.

From this site :

NOTES: The JDK's "bin" directory should be listed before "c:\\windows\\system32" and "c:\\windows" in the PATH. This is because some older Windows systems provide their own Java runtime (which is often out-dated) in these directories (try search for "java.exe" in your computer!).

Please update your PATH system ( not user) variable to include the location of your 32 bit Java 7 JDK. Both PATH and JAVA_HOME should be system variables.

It sounds like Windows may still be picking up on your 64 bit Java 7.

Answer 2

Update in enviromental variables JAVA_HOME as C:/program files/java/jdk version. And JRE_HOME as C:/Program files/java/jdk version/jre.