For Java applications, i.e. programs that are delivered (usually) as .jar files and started with java -jar xxx.jar or via a shortcut that does the same, the JRE that will be launched will be the first one found on the PATH.

If you installed a JRE or JDK, the likely places to find the .exes are below directories like C:\Program Files\JavaSoft\JRE\x.y.z. However, I've found some "out of the box" Windows installations to (also?) have copies of java.exe and javaw.exe in C:\winnt\system32 (NT and 2000) or C:\windows\system (Windows 95, 98). This is usually a pretty elderly version of Java: 1.3, maybe? You'll want to do java -version in a command window to check that you're not running some antiquated version of Java.

You can of course override the PATH setting or even do without it by explicitly stating the path to java.exe / javaw.exe in your command line or shortcut definition.

If you're running applets from the browser, or possibly also Java Web Start applications (they look like applications insofar as they have their own window, but you start them from the browser), the choice of JRE is determined by a set of registry settings:

Key: HKEY_LOCAL_MACHINE\Software\JavaSoft\Java Runtime EnvironmentName: CurrentVersionValue: (e.g.) 1.3

More registry keys are created using this scheme:

(e.g.) HKEY_LOCAL_MACHINE\Software\JavaSoft\Java Runtime Environment\1.3   HKEY_LOCAL_MACHINE\Software\JavaSoft\Java Runtime Environment\1.3.1

i.e. one for the major and one including the minor version number. Each of these keys has values like these (examples shown):

JavaHome    : C:\program Files\JavaSoft\JRE\1.3.1RuntimeLib  : C:\Program Files\JavaSoft\JRE\1.3.1\bin\hotspot\jvm.dllMicroVersion: 1

... and your browser will look to these settings to determine which JRE to fire up.

Since Java versions are changing pretty frequently, there's now a "wizard" called the "Java Control Panel" for manually switching your browser's Java version. This works for IE, Firefox and probably others like Opera and Chrome as well: It's the 'Java' applet in Windows' System Settings app. You get to pick any one of the installed JREs. I believe that wizard fiddles with those registry entries.

If you're like me and have "uninstalled" old Java versions by simply wiping out directories, you'll find these "ghosts" among the choices too; so make sure the JRE you choose corresponds to an intact Java installation!

Some other answers are recommending setting the environment variable JAVA_HOME. This is meanwhile outdated advice. Sun came to realize, around Java 2, that this environment setting is

1. unreliable, as users often set it incorrectly, and
2. unnecessary, as it's easy enough for the runtime to find the Java library directories, knowing they're in a fixed path relative to the path from which java.exe or javaw.exe was launched.

There's hardly any modern Java software left that needs or respects the JAVA_HOME environment variable.

More Information:

• http://java.sun.com/j2se/1.3/runtime_win32.html
• http://www.rgagnon.com/javadetails/java-0604.html

...and some useful information on multi-version support:

• http://www.rgagnon.com/javadetails/java-0604.html

I use to work on UNIX-like machines, but recently I have had to do some work with Java on a Windows 7 machine. I have had that problem and this is the I've solved it. It has worked right for me so I hope it can be used for whoever who may have this problem in the future.

These steps are exposed considering a default Java installation on drive C. You should change what it is necessary in case your installation is not a default one.

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_1.7mklink 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

Since Java 1.6, a java.exe is installed into %windir%\system32 that supports a "-version" command line option. You can use this to select a specific version to run, e.g.:

java -version:1.7 -jar [path to jar file]

will run a jar application in java 1.7, if it is installed.

See Oracle's documentation here: http://docs.oracle.com/javase/6/docs/technotes/tools/windows/java.html