Different Java version showing on command line

Question

I have recently checked on my Java version. I ran the command java -version and I found out that I was using java version 1.7.0_09. But when I tried to

Answer 1

In answer to the "actual" question:

Another thing that is weird is, I tried to check on environment variable settings, and it does not say anything about jdk1.7.0_09.

What happened here is that you installed jdk1.7.0_07 and then you auto-upgraded it. When that happens , it still uses the old folder name that you originally installed to.

After I install Java , I usually make a copy of the JDK directory and name it with the version number. Then, I can directly call a certain java like so:

@echo off
:: testjava.bat
set JAVA_HOME=C:\JDK1.x.xx
set PATH=%JAVA_HOME%\bin;%PATH%;.
java -version
pause

So, my recommendation is to set your JAVA_HOME system variable and PATH variable like I show above. This would override everything on your system so that your JDK of your choice is the default over the JRE.

Answer 2

That AppData path in your comment isn't on your path (supposedly, anyway), so that's probably not what it's using. Unfortunately, there isn't a which command on Windows either.

If you edit your path and move the C:\Program Files\Java\bin directory to the very beginning of the list and it still prints 1.7.0_09, then somehow you have JDK7u9 in your JDK7u7 folder. If not, browse to all the other directories on your path and open them 1-by-1 until you find the appropriate java file. Fortunately for you, your path is much shorter than mine.

Note that when doing:

> java -version

It may also look for java.bat and other non-exe extensions, so keep an eye out for this while you're searching your path. Try running:

> java.exe -version

That way you know you're looking for an exe file.

One last thing you can try:

> "C:\Program Files\Java\jdk1.7.0_07\bin\java" -version

If this returns 1.7.0_09, then something happened that updated your JDK in-place, which isn't supposed to happen, AFAIK (but I could be wrong).

Answer 3

It's possible to have many JRE side-by-side on a computer.

If the JRE is properly installed on Windows, informations about each version are stored in the registry. The installation process installs a special java.exe in the system PATH (%SYSTEMROOT%\System32). So you don't need to alter you PATH because this special java.exe will find the current JRE. From a command line, type java -version to display the current jre version installed.

With release 1.6, it's now possible to select a different JRE installation than the last one without any registry modification.

The JRE installation are listed in the registry in the key

HKEY_LOCAL_MACHINE\SOFTWARE\JavaSoft\Java Runtime Environment

Take this simple test class

public class ShowVersion {
 public static void main(String args[]) {
   System.out.println(System.getProperty("java.version"));
 }
}

On a system, with 1.6 and 1.5 installed. If you type

> java ShowVersion

It's probably the 1.6 JRE that will be used since it's the last installed.

To force the 1.5 JRE instead, use this command line.

> java -version:"1.5" ShowVersion

If the bytecode is incompatible with the given JRE then .. it won't work, of course.

ref : technote java 6

You can always give the complete path to use a specific installation. Launching the JVM this way does not use the registry setting at all.

>"C:\Program Files\Java\j2re1.4.1_02\bin\java" -version
java version "1.4.1_02"

source : Select a particular JRE from the command line

Answer 4

Adding the following will resolve your issue:

set JAVA_HOME="your jdk path"
set PATH=%JAVA_HOME%\bin;%PATH%.

Additionally if it does not work that means you have set the PATH for multiple java versions, include only the latest one and remove all from PATH variables.