java -version different than expected

Question

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?

Bozho · Accepted Answer · 2010-12-03 17:48:59Z

4

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

Bozho

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khachik · Accepted Answer · 2010-12-03 17:51:02Z

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

khachik

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3 Comments

Mark Peters Over a year ago

Bingo. It's invoking the one that resides in C:\Windows\System32

khachik Over a year ago

@Mark, I hate it... This forced to write which for windows.

Mark Peters Over a year ago

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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