Change Java default VM on Windows 7

Suppose we have installed Java 8 but for whatever reason we want to keep with Java 7.

1- Start a cmd as administrator

2- Go to C:\ProgramData\Oracle\Java

3- Rename the current directory javapath to javapath_<version_it_refers_to>. E.g.: rename javapath javapath_1.8

4- Create a javapath_<version_you_want_by_default> directory. E.g.: mkdir javapath_1.7

5- cd into it and create the following links:

cd javapath_1.7
mklink java.exe "C:\Program Files\Java\jre7\bin\java.exe"
mklink javaw.exe "C:\Program Files\Java\jre7\bin\javaw.exe"
mklink javaws.exe "C:\Program Files\Java\jre7\bin\javaws.exe"

6- cd out and create a directory link javapath pointing to the desired javapath. E.g.: mklink /D javapath javapath_1.7

7- Open the register and change the key HKEY_LOCAL_MACHINE\SOFTWARE\JavaSoft\Java Runtime Environment\CurrentVersion to have the value 1.7

At this point if you execute java -version you should see that you are using java version 1.7:

java version "1.7.0_71"
Java(TM) SE Runtime Environment (build 1.7.0_71-b14)
Java HotSpot(TM) 64-Bit Server VM (build 24.71-b01, mixed mode)

8- Finally it is a good idea to create the environment variable JAVA_HOME. To do that I create a directory link named CurrentVersion in C:\Program Files\Java pointing to the Java version I'm interested in. E.g.:

cd C:\Program Files\Java\
mklink /D CurrentVersion .\jdk1.7.0_71

9- And once this is done:

• Right click My Computer and select Properties.
• On the Advanced tab, select Environment Variables, and then edit/create JAVA_HOME to point to where the JDK software is located, in that case, C:\Program Files\Java\CurrentVersion