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Asked: 2026-05-23T15:57:52+00:00

I have set JAVA_HOME = C:\Program Files\Java\jdk1.6.0_26 in user defined variables and system variables

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I have set JAVA_HOME = C:\Program Files\Java\jdk1.6.0_26 in user defined variables and system variables and classpath = .; in user defined variables PATH = C:\Program Files\Java\jdk1.6.0_26\bin in system variables.

Now when I type java on cmd on windows 7 os then am getting message – Error could not open `C:\Program Files\Java\jre6\lib\i386\jvm.cfg'.

Thing to note here is that initially only jre6 was defined in Java folder and so I had to install jdk but still am getting this error any clue?

Update
I have able to get java information, all I did was initially I had set up path variable information in following some %SYSTEM_ROOT% and other variables in the PATH Variables values but now I just put C:\Program Files\Java\jdk1.6.0_26\bin as first value in PATH Variables value list and so it worked, guess %SYSTEM_ROOT% has pre-defined settings for Path which was pointing to old jre path location.

  1. Now when I run java -version, I
    get following information, how do i
    know if java installed is 32-bit or
    64-bit?

java version “1.6.0_26” Java(TM) SE
Runtime Environment (build
1.6.0_26-b03) Java HotSpot(TM) Client VM (build 20.1-b02, mixed mode,
sharing)

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1 Answer

  1. Editorial Team

    Your path variable is correct but for classpath variable use
    SET Classpath="C:\Program Files\Java\jdk1.6.0_26\lib"

    JVM uses class path at run time to search for .class files.

    By using “.;” JVM will search for .class files in bin directory whereas the compiled .class files of rt.jar are in lib directory.

    This why you’re facing errors.

    learn more about classpath variable @ http://programmingbulls.com/classpath-variable-java

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