java -version

Context:

JAVA_HOME= C:\Programmi\Java\jdk-1.5_16

If i launch the java -version under C:\Programmi\Java\jdk-1.5_16\bin
I have :
C:\Programmi\Java\jdk-1.5_16\bin>java -version
java version "1.5.0_15"
Java(TM) 2 Runtime Environment, Standard Edition (build 1.5.0_15-b04)
Java HotSpot(TM) Client VM (build 1.5.0_15-b04, mixed mode)

With the java -version command launched somewhere I have:
C:\Sonic\testConsegna\BARI\SONIC-BIN\MQ7.5>java -version
CompilerOracle: exclude progress/message/broker/AgentQueueDispatcher$DispatchThr
ead.threadMain
java version "1.6.0_20"
Java(TM) SE Runtime Environment (build 1.6.0_20-b02)
Java HotSpot(TM) Client VM (build 16.3-b01, mixed mode, sharing)

Why the result 1.6 if instead in the java_home I have set 1.5?
Thanks Menen

Because it actually looks in PATH and not JAVA_HOME, so I assume there's a reference to the 1.6 installation folder in PATH.

When you say java -version in windows , it first looks for any java executable in the current directory.

If it does not find one, then it goes for PATH

As you have mentioned, you have tried java -version in the bin folders of 2 different versions of java. So, obviously you will get the version of the java executable in that particular folder.

Try java -version in any other folder and check.

Also, this is what happens in Linux and UNIX too.

Sandeep Sanaboyina wrote:Also, this is what happens in Linux and UNIX too.

Are you sure about looking in the current directory as a default? I used not to be able to execute files in my current directory on Linux, until I added .: to my PATH.

Windows adds an implicit . at the start of the PATH variable, Linux (or better: almost all Linux shells like bash, csh, etc) doesn't.

Thank you, Rob. That explained it all.

Thanks I had to put the JAVA_HOME at the start of the path and now it does work fine
By Menen