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I'm having a problem with determining what version of java am I using:) cmd.exe is using java.exe from C:\WINDOWS\system32 as specified in Path environment variable.

C:\WINDOWS\system32>java.exe -version
java version "1.6.0_17"
Java(TM) SE Runtime Environment (build 1.6.0_17-b04)
Java HotSpot(TM) Client VM (build 14.3-b01, mixed mode)

But in windows when I right-click on that file and select Properties->Version->Full Version it says 1.6.0_22-b04.

Why?

asked Dec 3, 2010 at 17:46
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2 Answers 2

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This means you have two installations. Check your c:\program files\Java folder. Also check your PATH environment variable - it should point to the correct path.

answered Dec 3, 2010 at 17:48
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1

Change to a directory which is not in your PATH environment variable and run java -version. Note that the current directory always is the first entry in PATH.

answered Dec 3, 2010 at 17:51

3 Comments

Bingo. It's invoking the one that resides in C:\Windows\System32
@Mark, I hate it... This forced to write which for windows.
Yeah, the *nix alternative where you have to manually do ./someBinary is confusing at first ("why do I need to specify the directory if I'm already in it?") but avoids problems later.

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