CODEWITHSUNDEEP
javainheritancelog4j
Peter
Peter

Reputation: 123

log4j: How do I enable logging in a subclass for a method in a superclass

I have a logging statement in a method of my Superclass. I want to enable this statement only if the method is called for an Object of SubClassA.

public class SuperClass
{
private static Logger               logger  = Logger.getLogger(SuperClass.class);
public void test()
{
    logger.info("test...");
}
}

...

public class SubClassA extends SuperClass
{
private static Logger               logger  = Logger.getLogger(SubClassA.class);
}

...

public class SubClassB extends SuperClass
{
private static Logger               logger  = Logger.getLogger(SubClassB.class);
public static void main(String[] p_Args)
{
    SubClassA subClassA = new SubClassA();
    SubClassB subClassB = new SubClassB();
    subClassA.test();
    subClassB.test();
}
}

How do I enable logging in test() only for SubclassA?

log4j.logger.SuperClass=info //enables logging in the test() method for both Subclasses

log4j.logger.SubClassA=info //does nothing for the test() method

Upvotes: 4

Views: 5199

Answers (2)

Gerd Klima
Gerd Klima

Reputation: 1392

If you only want to enable logging in this one subclass you should define the logger (or an additional logger to the standard package/class named one) to a specific string (Logger.getLogger("SpecialSubClassALogger") and use it in SuperClass to log. You can then use log4j.logger.SpecialSubClassALogger=info.

Upvotes: 0

kan
kan

Reputation: 28951

I don't think it's possible, because logger doesn't know anything about classes and inheritance. A logger name is a simple textual name like "a.b.c.d". Maybe you could use subclass's class in super class, i.e. instead of:

private static Logger               logger  = Logger.getLogger(SuperClass.class);

use:

private Logger               logger  = Logger.getLogger(getClass());

or you could use both:

private Logger               subLogger  = Logger.getLogger(getClass());
private static Logger        logger  = Logger.getLogger(SuperClass.class);

and then you could use more sophisticated logic:

if(logger.isInfoEnabled() || subLogger.isInfoEnabled())
{
...
}

But if I were you, I would not use this magic, because logging should be as simple as possible (but not simpler).

Upvotes: 5

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