log4j

you need to specify debug level for the package that class belongs to

you don't need
" if ( log.isDebugEnabled() ){"

and perhaps you could try log4j.rootLogger=DEBUG

are you sure the properties file is on the classpath ? try a system.out after your  if ( log.isDebugEnabled() )


br,
themuppeteer

http://users.telenet.be/free_quiz

>>
you need to specify debug level for the package that class belongs to

How?

>>
and perhaps you could try log4j.rootLogger=DEBUG
i did and when it start tomcat i end up in an  infinite loop on my c: prompt tomcat console

Thanks

i use this is example(http://www.theserverside.com/discussions/thread.tss?thread_id=31659)
to create the property file below but i am still not getting any log output,... can someone help?

package actions;
...........
public class CreateCustomerAction extends AbstractAction {
	
    private static Log log = LogFactory.getLog(CreateCustomerAction.class);
.............
      if (log.isDebugEnabled()){
        	log.debug("just read firstName:"+ fn
        			+", lastName: " + ln 
        			+", emailAddress: "+ ad
        			+", custumerID: " + custId);
        }
    
the current log4j.properties file
=================================
log4j.logger.actions=DEBUG,CreateCustomerAction
 
llog4j.appender.CreateCustomerAction=org.apache.log4j.FileAppender
log4j.appender.CreateCustomerAction.File=createCustomerAction.log
log4j.appender.CreateCustomerAction.layout=org.apache.log4j.PatternLayout
log4j.appender.CreateCustomerAction.layout.ConversionPattern=%d{MM-dd@HH:mm:ss} %-5p (%13F:%L) %3x - %m%n
 
log4j.logger.org.apache.struts=DEBUG
org.hibernate.SQL=DEBUG
org.hibernate.type=DEBUG
org.hibernate.tool.hbm2ddl=DEBUG
org.hibernate.pretty=DEBUG
org.hibernate.cache=DEBUG
org.hibernate.transaction=DEBUG
org.hibernate.jdbc=DEBUG
org.hibernate.hql.ast.AST=DEBUG
org.hibernate.secure=DEBUG
org.hibernate=DEBUG

Select allOpen in new window

by adding this to your original properties file

log4j.logger.action=DEBUG, stdout

i added "log4j.logger.actions=DEBUG, stdout"  but still can't haave an output of what i added in my class.

The output generated is shown on screen shot added

Please see below the current log4J.properties.

log4j.rootLogger=WARN, stdout
log4j.logger.actions=DEBUG, stdout
 
log4j.appender.stdout=org.apache.log4j.ConsoleAppender
log4j.appender.stdout.layout=org.apache.log4j.PatternLayout
log4j.appender.stdout.layout.ConversionPattern=[%5p] %d{mm:ss}
(%F:%M:%L)%n%m%n%n
 
log4j.logger.org.apache.struts=DEBUG
org.hibernate.SQL=DEBUG
org.hibernate.type=DEBUG
org.hibernate.tool.hbm2ddl=DEBUG
org.hibernate.pretty=DEBUG
org.hibernate.cache=DEBUG
org.hibernate.transaction=DEBUG
org.hibernate.jdbc=DEBUG
org.hibernate.hql.ast.AST=DEBUG
org.hibernate.secure=DEBUG
org.hibernate=DEBUG

Select allOpen in new window

log4j.JPG

is WARN logging showing up from that class?

>> is WARN logging showing up from that class?
After i inserted these lines in that class, i have the screen shot below: note some WARN printed there. So, yes WARN is showing there.

           if (logger.isWarnEnabled()){
             logger.warn("just read firstName:"+ fn
                         +", lastName: " + ln
                         +", emailAddress: "+ ad
                         +", custumerID: " + custId);
      }


log4j.JPG